# Investigating Some Euler Sums

Common Topics: values, thus, middle, add, know

So, why only odd powers? Mostly because the even powers were solved by Leonard Euler in the 18th century. Since the “mathematical toolbox” at that time did not contain the required tools, he needed 6 years to prove the validity of his deductions. Now, however, we have much more powerful tools available, as I have shown in one of my previous insights (Using the Fourier Series to Find Some Interesting Sums).

Leaving the even powers aside, the odd powers are much more difficult. Using a computer, the values have been calculated to an awesome degree of precision, but as far as I know, no general closed-form expression has been found.

The first odd positive number is of course 1. The corresponding series is of course 1+1/2+1/3+⋯. This series has a special name – the harmonic series. Unfortunately, this sum diverges, albeit very slowly. It can be shown that the partial sum up to 1/N tends to ln⁡(N)+γ, where γ is the Euler-Mascheroni constant (approximately 0.5772). Therefore, we will start investigations at third powers and go from there.

Although no exact expression exists, some facts have been proven. The sum of the inverse cubes has been shown to be an irrational number (Apéry’s theorem).

## Stating the goal

What I expect to do in this insight: I will calculate the sums

• $\frac{1}{1^{p}}-\frac{1}{3^{p}}+\frac{1}{5^{p}}-\frac{1}{7^{p}}+\cdots$
• $\frac{1}{1^{p}}+\frac{1}{2^{p}}-\frac{1}{4^{p}}-\frac{1}{5^{p}}+\frac{1}{7^{p}} +\cdots$
• $\frac{1}{1^{p}}+\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{5^{p}}-\frac{1}{6^{p}}-\frac{1}{7^{p}}+\cdots$
• $\frac{1}{1^{p}}+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}}+\frac{1}{5^{p}}-\frac{1}{7^{p}}-\frac{1}{8^{p}}+\cdots$
• $\frac{1}{1^{p}}+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}}+\frac{1}{5^{p}}+\frac{1}{6^{p}}+\frac{1}{7^{p}}+\frac{1}{8^{p}}+\frac{1}{9^{p}}+\frac{1}{10^{p}}+\frac{1}{11^{p}}-\frac{1}{13^{p}}-\frac{1}{14^{p}}-\cdots$

for p = 3, 5, 7 and 9. The method works just as well if p is an odd integer >9, but the numbers used in the arithmetic get very unwieldy.

I will also discuss some extension to these sums.

#### Remark

In this paper, the fact that $e^{in\pi}=e^{-in\pi}=-1$ is going to be used repeatedly without any explanation.

Also, there will be instances where some trivial details are skipped. This insight is long enough as it is.

### Lemma 1

Let $g(x)=e^{ikx}x^{2m}$ where k and m are integers. Then $g(π)-g(-π)=0$.

#### Proof:

$g(\pi)-g(-\pi)=(-1)^{k}\pi^{2m}-(-1)^{k}(- \pi)^{2m}=(-1)^{k}((\pi^{2})^{m}-((-\pi)^{2})^{m})=0$

### Lemma 2

Let $g(x)=e^{ikx}\sum_{m=1}^{N}a_{m}x^{2m}$ where N is a finite integer. Then $g(π)-g(-π)=0$.

#### Proof:

Repeated application of Lemma 1.

### Lemma 3

Let p be an odd integer (p=2N+1). Then there exists a polynomial h(x) such that

1. $h(\pi)-h(-\pi)=0$
2. $h'(\pi)-h'(-\pi)=0$
3. $h^{(q)}(\pi)-h^{(q)}(-\pi)=0$ for all q-derivatives where q<p.

#### Proof

Let $a_{1}\pi^{p-1}x+a_{3}\pi^{p-3}x^{3}+…-x^{p}$ (only odd powers of x). Our requirements give the following set of equations:

1. $a_{1}+a_{3}+\cdots =1$
2. $6a_{3}+20a_{5}+\cdots=p(p-1)$

This ends up in a neat solution for all $a_{k}$. Observe that the odd derivatives of h(x) are sums of even powers of x and Lemma 2 assures that the corresponding equations are satisfied.

### Actual values for h(x):

• $h_{3}(x)=-x^{3}+\pi^{2}x$
• $h_{5}(x)=-x^{5}+\frac{10}{3}\pi^{2}x^{3}-\frac{7}{3}\pi^{4}x$
• $h_{7}(x)=-x^{7}+7\pi^{2}x^{5}- \frac{49}{3}\pi^{4}x^{3}+\frac{31}{3}\pi^{6}x$
• $h_{9}(x)=-x^{9}+12\pi^{2}x^{7}- \frac{294}{5}\pi^{4}x^{5}+124\pi^{6}x^{3}- \frac{381}{5}\pi^{8}x$

#### Remark

The h(x) could just as easily be found requiring the coefficient of x to be 1. This results in the following expressions:

• $g_{3}(x)=x-\frac{1}{\pi^{2}}x^{3}$
• $g_{5}(x)=x-\frac{10}{7\pi^{2}}x^{3}+\frac{3}{7\pi^{4}}x^{5}$
• $g_{7}(x)=x-\frac{49}{31\pi^{2}}x^{3}+\frac{21}{31\pi^{4}}x^{5}-\frac{3}{31\pi^{6}}x^{7}$

These expressions bear a close resemblance to the truncated series for sin(x) In fact, they are a better fit in the interval [-π, π] than the truncated series.

The Fourier series of a repetitive waveform is usually calculated in order to show the harmonic distortion of the waveform. What we really care about to do, is calculating the harmonic distortion of our polynomials h(x).

### Lemma 4

Let h(x) be as in Lemma 3. Then the Fourier transform coefficients of h(x) is given by $a_{0}=0, a_{n}=(-1)^{n}\frac{p!}{(in)^{p}}$.

Thus, we have $h(x)=(-1)^{n}(\sum_{n=-\infty}^{-1}\frac{p!}{(in)^{p}}+\sum_{1}^{\infty}\frac{p!}{(in)^{p}})$. Combine the two sums: $h(x)=\frac{p!}{i^{p}}\sum_{n=1}^{\infty}(-1)^{n}\frac{e^{inx}-e^{-inx}}{n^{p}}=(-1)^{\frac{p+1}{2}}\sum_{n=1}^{\infty}\frac{e^{in(\pi-x)}-e^{-in(\pi-x)}}{in^{p}}=(-1)^{\frac{p+1}{2}}\cdot 2\cdot p! \sum_{n=1}^{\infty}\frac{\sin (n(\pi -x))}{n^{p}}$.

#### Remark

The Fourier series of a real-valued function is usually calculated using $\sin^{n}(x)$ and $\cos^{n}(x)$ and since we are dealing with odd functions (only odd powers of x), we know that the result will only contain powers of sin(x). Using complex expressions is just a personal preference – and it ends up with the same results.

Someone may wonder if the Fourier sums converge to the value of the original expression. This is trivially fulfilled since polynomials have continuous derivatives everywhere.

One problem with using the Fourier series is that the resulting coefficients contain $\sin(\frac{k\pi}{M})$ for integer values of k. Thus, whenever $(2r+1)M\leq k < (2r+2)M$ (r an integer), $\sin(\frac{k\pi}{M})$ will be negative.

With these cautions out of the way, let us start with the results!

### 1. Actual values for x=π/2

The values of the sums in this chapter are:

• $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{3}}=\frac{\pi^{3}}{32}$
• $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{5}}=\frac{5\pi^{5}}{1536}$
• $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{7}}=\frac{61\pi^{7}}{184320}$
• $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{9}}=\frac{277\pi^{9}}{8257536}$

#### Proofs

The values for $\sin(\frac{n\pi}{2})$ are (1, 0, -1, 0). Also $\pi-\frac{\pi}{2}=\frac{\pi}{2}$. Thus:

$h_{3}(\frac{\pi}{2})=\frac{3}{8}\pi^{3}=12\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{3}}$, thus $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{3}}=\frac{\pi^{3}}{32}$.

$h_{5}(\frac{\pi}{2})=\frac{-25}{32}\pi^{5}=-240\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{5}}$, thus $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{5}}=\frac{5\pi^{5}}{243}$

$h_{7}(\frac{\pi}{2})=\frac{427}{128}\pi^{7}=10080\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{7}}$, thus $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{7}}=\frac{61\pi^{7}}{184320}$

$h_{9}(\frac{\pi}{2})=\frac{-12465}{512}\pi^{9}=-725760\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{9}}$, thus $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{9}}=\frac{277\pi^{9}}{8257536}$

### 2. Actual values for x=2π/3

The values of the sums in this chapter are:

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(3n+1)^{3}}+\frac{1}{(3n+2)^{3}})=\frac{5\sqrt{3}}{243}\pi^{3}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(3n+1)^{5}}+\frac{1}{(3n+2)^{5}})=\frac{17\sqrt{3}}{8748}\pi^{5}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(3n+1)^{7}}+\frac{1}{(3n+2)^{7}})=\frac{91\sqrt{3}}{472392}\pi^{7}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(3n+1)^{9}}+\frac{1}{(3n+2)^{9}})=\frac{207913\sqrt{3}}{10713850560}\pi^{9}$

#### Proofs

The values for $\sin(\frac{n\pi}{3})$ are ($\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},0, -\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2},0$. Also $\pi-\frac{\pi}{3}=\frac{2\pi}{3}$. Thus:

• $h_{3}(\frac{2\pi}{3})=\frac{10}{27}\pi^{3}=12\sum_{n=0}^{\infty}(\frac{\frac{\sqrt{3}}{2}}{((3n+1)^{3}}+\frac{\frac{\sqrt{3}}{2}}{((3n+2)^{3}})$, thus $\sum_{n=0}^{\infty}(\frac{1}{((3n+1)^{3}}+\frac{1}{((3n+2)^{3}})=\frac{5\sqrt{3}}{243}\pi^{3}$
• $h_{5}(\frac{2\pi}{3})=\frac{-170}{243}\pi^{5}=-240\sum_{n=0}^{\infty}(\frac{\frac{\sqrt{3}}{2}}{((3n+1)^{5}}+\frac{\frac{\sqrt{3}}{2}}{((3n+2)^{5}})$, thus $\sum_{n=0}^{\infty}(\frac{1}{((3n+1)^{5}}+\frac{1}{((3n+2)^{5}})=\frac{17\sqrt{3}}{8748}\pi^{5}$
• $h_{7}(\frac{2\pi}{3})=\frac{6870}{2187}\pi^{7}=10080\sum_{n=0}^{\infty}(\frac{\frac{\sqrt{3}}{2}}{((3n+1)^{7}}+\frac{\frac{\sqrt{3}}{2}}{((3n+2)^{7}})$, thus $\sum_{n=0}^{\infty}(\frac{1}{((3n+1)^{7}}+\frac{1}{((3n+2)^{7}})=\frac{91\sqrt{3}}{472392}\pi^{7}$
• $h_{9}(\frac{2\pi}{3})=\frac{-415826}{19683}\pi^{9}=725760\sum_{n=0}^{\infty}(\frac{\frac{\sqrt{3}}{2}}{((3n+1)^{9}}+\frac{\frac{\sqrt{3}}{2}}{((3n+2)^{9}})$, thus $\sum_{n=0}^{\infty}(\frac{1}{((3n+1)^{9}}+\frac{1}{((3n+2)^{9}})=\frac{207913\sqrt{3}}{10713850560}\pi^{9}$

### 3. Actual values for x=3π/4

The values of the sums in this chapter are:

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}+\frac{1}{(4n+2)^{3}}+\frac{1}{(4n+3)^{3}})=\frac{1+6\sqrt{2}}{256}\pi^{3}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}+\frac{1}{(4n+2)^{5}}+\frac{1}{(4n+3)^{5}})=\frac{5+114\sqrt{2}}{49152}\pi^{5}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}+\frac{1}{(4n+2)^{7}}+\frac{1}{(4n+3)^{7}})=\frac{427+38682\sqrt{2}}{165150720}\pi^{7}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}+\frac{1}{(4n+2)^{9}}+\frac{1}{(4n+3)^{9}})=(\frac{277}{4227858432}+\frac{83579\sqrt{2}}{3523215360})\pi^{9}$

#### Proofs

The values for $\sin(\frac{n\pi}{4})$ are $(\frac{\sqrt{2}}{2}, 1, \frac{\sqrt{2}}{2}, 0, -\frac{\sqrt{2}}{2}, -1, -\frac{\sqrt{2}}{2},0)$. Also $\pi-\frac{\pi}{4}=\frac{3\pi}{4}$. Thus:

##### Third powers

$h_{3}(\frac{3\pi}{4})=\frac{21}{64}\pi^{3}=12\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{3}}+\frac{1}{(4n+2)^{3}}+\frac{\frac{\sqrt{2}}{2}}{(4n+3)^{3}})$.

The middle element we already know from $h_{3}(\frac{\pi}{2}):\frac{\pi^{3}}{32\cdot 2^{3}}=\frac{\pi^{3}}{256}$. Extract it: $\frac{21}{64}\pi^{3}-\frac{12}{256}\pi^{3}=\frac{18}{64}\pi^{3}$.

Thus $\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{3}}+\frac{\frac{\sqrt{2}}{2}}{(4n+3)^{3}})=\frac{18}{768}\pi^{3}$. Divide out the square root: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}+\frac{1}{(4n+3)^{3}})=\frac{18\sqrt{2}}{768}\pi^{3}$. Add back the middle element: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}+\frac{1}{(4n+2)^{3}}+\frac{1}{(4n+3)^{3}})=\frac{3+18\sqrt{2}}{768}\pi^{3}=\frac{1+6\sqrt{2}}{256}\pi^{3}$.

##### Fifth powers

$h_{5}(\frac{3\pi}{4})=\frac{-595}{1024}\pi^{5}=-240\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{5}}+\frac{1}{(4n+2)^{5}}+\frac{\frac{\sqrt{2}}{2}}{(4n+3)^{5}})$. Change sign: $\frac{595}{1024}\pi^{5}=240\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{5}}+\frac{1}{(4n+2)^{5}}+\frac{\frac{\sqrt{2}}{2}}{(4n+3)^{5}})$. The middle element we already know from $h_{5}(\frac{\pi}{2}):\frac{5\pi^{5}}{1536\cdot 2^{3}}=\frac{5\pi^{3}}{49152}$. Extract it: $\frac{119}{49152}\pi^{5}-\frac{5}{49152}\pi^{5}=\frac{114}{49152}\pi^{5}$. Thus: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{5}}+\frac{\frac{\sqrt{2}}{2}}{(4n+3)^{5}})=\frac{114}{49152}\pi^{5}=\frac{19}{8192}\pi^{5}$. Divide out the square root: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}+\frac{1}{(4n+3)^{5}})=\frac{19\sqrt{2}}{8192}\pi^{5}$. Add back the middle element: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}+\frac{1}{(4n+2)^{5}}+\frac{1}{(4n+3)^{5}})=(\frac{114\sqrt{2}}{49152}+\frac{5}{49152})\pi^{5}=\frac{5+114\sqrt{2}}{49152}\pi^{5}$.

##### Seventh powers

$h_{7}(\frac{3\pi}{4})=\frac{39109}{16384}\pi^{7}=10080\sum_{n=0}^{\infty}(-1)^{n}(\frac{\sqrt{2}}{2(4n+1)^{7}}+\frac{1}{(4n+2)^{7}}+\frac{\sqrt{2}}{2(4n+3)^{7}})$. The middle element we already know from $h_{7}(\frac{\pi}{2}):\frac{61}{184320}\pi^{7}\cdot\frac{1}{128}$. Extract it: $\frac{39109}{16384}\pi^{7}-\frac{61\cdot 10080}{184320\cdot 128}\pi^{7}=\frac{5631690-614880}{23592960}\pi^{7}=\frac{19341}{8192}\pi^{7}$. Thus $10080\sum_{n=0}^{\infty}(-1)^{n}(\frac{\sqrt{2}}{2(4n+1)^{7}}+\frac{\sqrt{2}}{2(4n+3)^{7}})=\frac{19341}{8192}\pi^{7}$ or $\sum_{n=0}^{\infty}(-1)^{n}(\frac{\sqrt{2}}{2(4n+1)^{7}}+\frac{\sqrt{2}}{2(4n+3)^{7}})=\frac{19341}{8192\cdot 10080}\pi^{7}$. Divide out the square root: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}+\frac{1}{(4n+3)^{7}})=\frac{19341\sqrt{2}}{8257360}\pi^{7}$. Add back the middle element: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}+\frac{1}{(4n+2)^{7}}+\frac{1}{(4n+3)^{7}})=\frac{427+38682\sqrt{2}}{165150720}\pi^{7}$.

##### Ninth powers

$h_{9}(\frac{3\pi}{4})=\frac{-4525731}{262144}\pi^{9}=-725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{\sqrt{2}}{2(4n+1)^{9}}+\frac{1}{(4n+2)^{9}}+\frac{\sqrt{2}}{2(4n+3)^{9}})$ . Change sign: $\frac{4525731}{262144}\pi^{9}=725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{\sqrt{2}}{2(4n+1)^{9}}+\frac{1}{(4n+2)^{9}}+\frac{\sqrt{2}}{2(4n+3)^{9}})$. The middle element we already know from $h_{9}(\frac{\pi}{2}):\frac{277}{8257536}\pi^{9}\cdot\frac{1}{512}$. Extract it: $\frac{4525731}{262144}\pi^{9}-\frac{277\cdot 725760}{8257536\cdot 512}\pi^{9}=\frac{2256633}{131072}\pi^{9}$. Thus: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{\sqrt{2}}{2(4n+1)^{9}}+\frac{\sqrt{2}}{2(4n+3)^{9}})=\frac{2256633}{131072\cdot 725760}\pi^{9}$. Divide out the square root: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}+\frac{1}{(4n+3)^{9}})=\frac{2256633\sqrt{2}}{131072\cdot 725760}\pi^{9}$. Add back the middle element: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}+\frac{1}{(4n+2)^{9}}+\frac{1}{(4n+3)^{9}})=(\frac{277}{4227858432}+\frac{83579\sqrt{2}}{3523215360})\pi^{9}$.

### 4. Actual values for x=5π/6

The values of the sums in this chapter are:

$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{3}}+\frac{1}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}+\frac{1}{(6n+4)^{3}}+\frac{1}{(6n+5)^{3}})=\frac{261+20\sqrt{3}}{7776}\pi^{3}$

$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}+\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}+\frac{1}{(6n+4)^{5}}+\frac{1}{(6n+5)^{5}})=(\frac{1225}{373248}+\frac{17\sqrt{3}}{279936})\pi^{5}$

$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{7}}+\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}+\frac{1}{(6n+4)^{7}}+\frac{1}{(6n+5)^{7}})=(\frac{133529}{403107840}+\frac{91\sqrt{3}}{60466176})\pi^{7}$

$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{9}}+\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}+\frac{1}{(6n+4)^{9}}+\frac{1}{(6n+5)^{9}})=(\frac{1090549}{32506616218}+\frac{207913\sqrt{3}}{5485491486720)})\pi^{9}$

#### Proofs

The values for $\sin(\frac{n\pi}{6})$ are $\frac{1}{2}, \frac{\sqrt{3}}{2},1, \frac{\sqrt{3}}{2},\frac{1}{2}, 0, -\frac{1}{2}, -\frac{\sqrt{3}}{2},-1, -\frac{\sqrt{3}}{2},-\frac{1}{2}, 0,$. Also $\pi-\frac{\pi}{6}=\frac{5\pi}{6}$. Thus:

##### Third powers

$h_{3}(\frac{5\pi}{6})=\frac{55}{216}\pi^{3}=12\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{3}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{3}}+\frac{1}{2}\frac{1}{(6n+5)^{3}})$. The middle element we already know from $h_{3}(\frac{\pi}{2}):\frac{\pi^{3}}{32\cdot 3^{3}}=\frac{\pi^{3}}{864}$. Extract it: $\frac{55}{216}\pi^{3}-\frac{12}{864}\pi^{3}=\frac{13}{54}\pi^{3}$.

The sum of the two elements close to the middle we already know from $h_{3}(\frac{2\pi}{3}): \frac{\sqrt{3}}{2}\cdot\frac{12\cdot 5\sqrt{3}}{243\cdot 2^{3}}\pi^{3}$. Extract them: $\frac{13}{54}\pi^{3}-\frac{6\cdot 5\cdot 3}{243\cdot 2^{3}}\pi^{3}=\frac{7}{36}\pi^{3}$. Thus $12\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{3}}+\frac{1}{2}\frac{1}{(6n+5)^{3}})=\frac{7}{36}\pi^{3}$. Take the ½ outside:  $12\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{3}}+\frac{1}{(6n+5)^{3}})=\frac{7}{18}\pi^{3}$. Now we can add them all up: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{3}}+\frac{1}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}+\frac{1}{(6n+4)^{3}}+\frac{1}{(6n+5)^{3}})=(\frac{7}{18\cdot 12}+\frac{5\sqrt{3}}{243\cdot 8}+\frac{1}{864})\pi^{3}=\frac{261+20\sqrt{3}}{7776}\pi^{3}$.

##### Fifth powers

$h_{5}(\frac{5\pi}{6})=\frac{-3245}{7776}\pi^{5}=-240\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{5}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{5}}+\frac{1}{2}\frac{1}{(6n+5)^{5}})$. Change sign: $\frac{3245}{7776}\pi^{3}=240\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{5}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{5}}+\frac{1}{2}\frac{1}{(6n+5)^{5}})$. The middle element we already know from $h_{5}(\frac{\pi}{2}):\frac{5\pi^{3}}{1536\cdot 3^{5}}=\frac{5\pi^{5}}{373248}$. Extract it: $\frac{3245}{7776}\pi^{5}-\frac{5\cdot 240}{373248}\pi^{5}=\frac{805}{1944}\pi^{5}$. The two elements close to the middle we already know from $h_{5}(\frac{2\pi}{3}): \frac{\sqrt{3}}{2}\cdot\frac{17\sqrt{3}}{8748\cdot 2^{5}}\pi^{5}=\frac{17}{18664}\pi^{5}$. Extract them: $\frac{805}{1944}\pi^{5}-\frac{17}{18664}\pi^{5}=\frac{1525}{3888}\pi^{5}$. Thus $240\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2(6n+1)^{5}}+\frac{1}{2(6n+5)^{5}})=\frac{1525}{3888}\pi^{5}$. Take the ½ outside: $120\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}+\frac{1}{(6n+5)^{5}})=\frac{1525}{3888}\pi^{5}$. Now we can add them all up: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}+\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}+\frac{1}{(6n+4)^{5}}+\frac{1}{(6n+5)^{5}})=(\frac{5}{373248}\pi^{5}+\frac{17\sqrt{3}}{279936}\pi^{5}+\frac{1525}{3888\cdot 120})\pi^{5}$ $=(\frac{1225}{373248}+\frac{17\sqrt{3}}{279936})\pi^{5}$.

##### Seventh powers

$h_{7}(\frac{5\pi}{6})=\frac{473935}{279936}\pi^{7}=10080\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{7}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{7}}+\frac{1}{2}\frac{1}{(6n+5)^{7}})$. The middle element we already know from $h_{7}(\frac{\pi}{2}):\frac{61\pi^{7}}{184320\cdot 3^{7}}$. Extract it: $\frac{473935}{279936}\pi^{7}-\frac{61\cdot 10080}{184320\cdot 2157}\pi^{7}=\frac{1353}{7776}\pi^{7}$. The sum of the two elements close to the middle we already know from $h_{7}(\frac{2\pi}{3}): \frac{\sqrt{3}}{2}\cdot\frac{91\sqrt{3}}{472392\cdot 2^{7}}\pi^{7}=\frac{91}{40310784}\pi^{7}$. Extract them: $\frac{13153}{7776}\pi^{7}-\frac{91\cdot 10080}{40310784}\pi^{7}=\frac{233569}{139968}\pi^{7}$. Thus $10080\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2(6n+1)^{7}}+\frac{1}{2(6n+5)^{7}})=\frac{233569}{139968}\pi^{7}$. Take the ½ outside: $5040\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{7}}+\frac{1}{(6n+5)^{7}})=\frac{233569}{139968}\pi^{7}$. Now we can add them all up: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{7}}+\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}+\frac{1}{(6n+4)^{7}}+\frac{1}{(6n+5)^{7}})=\\(\frac{61}{403107840}+\frac{91\sqrt{3}}{60466176}+\frac{233569}{139968\cdot 5040})\pi^{7}$ $=(\frac{133529}{403107840}+\frac{91\sqrt{3}}{60466176})\pi^{7}$.

##### Ninth powers

$h_{9}(\frac{5\pi}{6})=\frac{-123108821}{10077696}\pi^{9}=-725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{9}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{9}}+\frac{1}{2}\frac{1}{(6n+5)^{9}})$. Change sign: $\frac{123108821}{10077696}\pi^{9}=725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{9}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{9}}+\frac{1}{2}\frac{1}{(6n+5)^{9}})$. The middle element we already know from $h_{9}(\frac{\pi}{2}):\frac{277\pi^{9}}{8257536\cdot 3^{9}}$ . Extract it: $\frac{123108821}{10077696}\pi^{9}-\frac{277\cdot 725760}{8257536\cdot 19683}\pi^{9}=\frac{30774089}{2519424}\pi^{9}$.

The sum of the two elements close to the middle we already know from $h_{9}(\frac{2\pi}{3}): \frac{\sqrt{3}}{2}\cdot\frac{207913\sqrt{3}}{10713850560\cdot 2^{9}}\pi^{9}=\frac{207913}{7142567040\cdot 512}\pi^{9}$. Extract them: $\frac{30774089}{2519424}\pi^{9}-\frac{207913\cdot 725760}{7142567040\cdot 512}\pi^{9}=\frac{47709095}{3919104}\pi^{9}$. Thus $725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2(6n+1)^{9}}+\frac{1}{2(6n+5)^{9}})=\frac{47709095}{3919104}\pi^{9}$. Take the ½ outside: $362880\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2(6n+1)^{9}}+\frac{1}{2(6n+5)^{9}})=\frac{47709095}{3919104}\pi^{9}$. Now we can add them all up: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{9}}+\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}+\frac{1}{(6n+4)^{9}}+\frac{1}{(6n+5)^{9}})=(\frac{277}{8257536\cdot 3^{9}}+\frac{207913\sqrt{3}}{10713850560\cdot 2^{9}}+\frac{47709095}{3919104\cdot 362880})\pi^{9}$ $=(\frac{1090549}{32506616218}+\frac{207913\sqrt{3}}{5485491486720})\pi^{9}$.

### 5. Actual values for x=11π/12

Figuring out how to set up the calculations for this chapter took almost a year and when I found a way, the large number of digits made it almost impossible to verify the results. Fortunately someone at Physics Forum pointed me to an online calculator that could handle 20-digit numbers and more, and also reduce the fractions as much as possible.

The values of the sums in this chapter are:

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+2)^{3}}+\frac{1}{(12n+3)^{3}}+\frac{1}{(12n+4)^{3}}+\frac{1}{(12n+5)^{3}}+ \frac{1}{(12n+6)^{3}}+\frac{1}{(12n+7)^{3}}$$+\frac{1}{(12n+8)^{3}}+\frac{1}{(12n+9)^{3}}+\frac{1}{(12n+10)^{3}}+\frac{1}{(12n+11)^{3}})=(\frac{29}{6912}+\frac{\sqrt{2}}{1152}+\frac{5\sqrt{3}}{15552}+\frac{23\sqrt{6}}{1728})\pi^{3}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+2)^{5}}+\frac{1}{(12n+3)^{5}}+\frac{1}{(12n+4)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+8)^{5}}$$+\frac{1}{(12n+9)^{5}}+\frac{1}{(12n+10)^{5}}+\frac{1}{(12n+11)^{5}})=(\frac{1225}{11943936}+\frac{19\sqrt{2}}{1990656}+\frac{17\sqrt{3}}{8957952}+\frac{3985\sqrt{6}}{2985984})\pi^{5}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+2)^{7}}+\frac{1}{(12n+3)^{7}}+\frac{1}{(12n+4)^{7}}+\frac{1}{(12n+5)^{7}}+\frac{1}{(12n+6)^{7}}+\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+8)^{7}}$$+\frac{1}{(12n+9)^{7}}+\frac{1}{(12n+10)^{7}}+\frac{1}{(12n+11)^{7}})=(\frac{133529}{51597803520}+\frac{307\sqrt{2}}{2866544640}+\frac{91\sqrt{3}}{7739670528}+\frac{1743623\sqrt{6}}{12899450880})\pi^{7}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+2)^{9}}+\frac{1}{(12n+3)^{9}}+\frac{1}{(12n+4)^{9}}+\frac{1}{(12n+5)^{9}}+\frac{1}{(12n+6)^{9}}+\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+8)^{9}}$$+\frac{1}{(12n+9)^{9}}+\frac{1}{(12n+10)^{9}}+\frac{1}{(12n+11)^{9}})=(\frac{1090549}{16643387503616}+\frac{83579\sqrt{2}}{69347447930880}+\frac{207913\sqrt{3}}{15552}+\frac{202681253\sqrt{6}}{10402117189632})\pi^{9}$

#### Proofs

The values for $\sin(\frac{\pi}{12})$ are ($a, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, b, 1, b, \frac{\sqrt{3}}{2}, \frac{\sqrt{2}}{2}, \frac{1}{2}, a, 0$), where  $a=\sin(\frac{\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4}$ and  $b=\sin(\frac{5\pi}{12})=\frac{\sqrt{6}+\sqrt{2}}{4}$. Of course  $\pi-\frac{\pi}{12}=\frac{11\pi}{12}$. Thus:

##### Third powers

$h_{3}(\frac{11\pi}{12})=\frac{253}{1728}\pi^{3}=12\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{3}}+\frac{1}{2(12n+2)^{3}}+\frac{\sqrt{2}}{2(12n+3)^{3}}+\frac{\sqrt{3}}{2(12n+4)^{3}}+\frac{b}{(12n+5)^{3}}+\frac{1}{(12n+6)^{3}}+\frac{b}{(12n+7)^{3}}$ $+\frac{\sqrt{3}}{2(12n+8)^{3}}+\frac{\sqrt{2}}{2(12n+9)^{3}}+\frac{1}{2(12n+10)^{3}}+\frac{a}{(12n+11)^{3}})$, which means that $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{3}}+\frac{1}{2(12n+2)^{3}}+\frac{\sqrt{2}}{2(12n+3)^{3}}+\frac{\sqrt{3}}{2(12n+4)^{3}}+\frac{b}{(12n+5)^{3}}+\frac{1}{(12n+6)^{3}}+\frac{b}{(12n+7)^{3}}+\frac{\sqrt{3}}{2(12n+8)^{3}}$$+\frac{\sqrt{2}}{2(12n+9)^{3}}+\frac{1}{2(12n+10)^{3}}+\frac{a}{(12n+11)^{3}})=\frac{253}{20736}\pi^{3}$. From (4) above we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{3}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{3}}+\frac{1}{2}\frac{1}{(6n+5)^{3}})=\frac{55}{292}\pi^{3}$. Divide by $2^{3}$ and subtract: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{3}}+\frac{\sqrt{2}}{2(12n+3)^{3}}+\frac{b}{(12n+5)^{3}}+\frac{b}{(12n+7)^{3}}+\frac{\sqrt{2}}{2(12n+9)^{3}}+\frac{a}{(12n+11)^{3}})=\frac{253}{20736}\pi^{3}-\frac{55}{20736}\pi^{3}=\frac{11}{1152}\pi^{3}$.

Now split the resultant expression:

$\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{3}}+\frac{\sqrt{2}}{2(12n+3)^{3}}+\frac{b}{(12n+5)^{3}}+\frac{b}{(12n+7)^{3}}+\frac{\sqrt{2}}{2(12n+9)^{3}}+\frac{a}{(12n+11)^{3}})$$=\frac{\sqrt{6}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+5)^{3}}+\frac{1}{(12n+7)^{3}}+\frac{1}{(12n+11)^{3}})$$-\frac{\sqrt{2}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+3)^{3}}-\frac{1}{(12n+5)^{3}}-\frac{1}{(12n+7)^{3}}+\frac{1}{(12n+9)^{3}}+\frac{1}{(12n+11)^{3}})$$+ \frac{3\sqrt{2}}{4} \sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+3)^{3}}+\frac{1}{(12n+9)^{3}})$.

Let $S_{1}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+5)^{3}}+\frac{1}{(12n+7)^{3}}+\frac{1}{(12n+11)^{3}})$, $S_{2}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+3)^{3}}-\frac{1}{(12n+5)^{3}}-\frac{1}{(12n+7)^{3}}+\frac{1}{(12n+9)^{3}}+\frac{1}{(12n+11)^{3}})$ and $S_{3}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+3)^{3}}+\frac{1}{(12n+9)^{3}})$. This means that $\frac{\sqrt{6}}{4}S_{1}-\frac{\sqrt{2}}{4}S_{2}+\frac{3\sqrt{2}}{4}S_{3}=\frac{11}{1152}\pi^{3}$

From (3) above we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}+\frac{1}{(4n+3)^{3}})=\frac{3\sqrt{2}}{128}\pi^{3}$. If we divide by $3^{3}$ we get $S_{3}=\frac{\sqrt{2}}{1152}\pi^{3}$. If we instead expand the formula, we get $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}+\frac{1}{(4n+3)^{3}})=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+3)^{3}}-\frac{1}{(12n+5)^{3}}-\frac{1}{(12n+7)^{3}}+\frac{1}{(12n+9)^{3}}+\frac{1}{(12n+11)^{3}})$, which is equal to $S_{2}$. Since $\frac{\sqrt{6}}{4}S_{1}=\frac{11}{1152}\pi^{3}+\frac{\sqrt{2}}{4}(S_{2}-3S_{3})=\frac{11}{1152}\pi^{3}+\frac{\sqrt{2}}{4}(\frac{27\sqrt{2}}{1152}-\frac{3\sqrt{2}}{1152})\pi^{3}=\frac{12+11}{1152}\pi^{3}=\frac{23}{1153}\pi^{3}$ , we get $S_{1}=\frac{4}{\sqrt{6}}\cdot\frac{23}{1153}\pi^{3}=\frac{\sqrt{6}}{6}\cdot \frac{92}{1152}\pi^{3}=\frac{23\sqrt{6}}{1728}\pi^{3}$. Add in $S_{3}:\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+3)^{3}}+\frac{1}{(12n+5)^{3}}+\frac{1}{(12n+7)^{3}}+\frac{1}{(12n+9)^{3}}+\frac{1}{(12n+11)^{3}})=\frac{23\sqrt{6}}{1728}\pi^{3}$. Now the sum from (4): $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{3}}+\frac{1}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}+\frac{1}{(6n+4)^{3}}+\frac{1}{(6n+5)^{3}})=\frac{261+20\sqrt{3}}{7776}\pi^{3}$, which means that $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+2)^{3}}+\frac{1}{(12n+4)^{3}}+\frac{1}{(12n+6)^{3}}+\frac{1}{(12n+8)^{3}}+\frac{1}{(12n+10)^{3}})=\frac{261+20\sqrt{3}}{62208}\pi^{3}$. Add it:$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+2)^{3}}+\frac{1}{(12n+3)^{3}}+\frac{1}{(12n+4)^{3}}+\frac{1}{(12n+5)^{3}}+ \frac{1}{(12n+6)^{3}}+\frac{1}{(12n+7)^{3}}+\frac{1}{(12n+8)^{3}}$ $+\frac{1}{(12n+9)^{3}}+\frac{1}{(12n+10)^{3}}+\frac{1}{(12n+11)^{3}})=\pi^{3}(\frac{23\sqrt{6}}{1728}+\frac{\sqrt{2}}{1152}+\frac{261+20\sqrt{3}}{62208})=(\frac{29}{6912}+\frac{\sqrt{2}}{1152}+\frac{5\sqrt{3}}{15552}+\frac{23\sqrt{6}}{1728})\pi^{3}$ .

##### Fifth powers

$h_{5}(\frac{11\pi}{12})=\frac{-54395}{248832}\pi^{5}=-240\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{5}}+\frac{1}{2(12n+2)^{5}}+\frac{\sqrt{2}}{2(12n+3)^{5}}+\frac{\sqrt{3}}{2(12n+4)^{5}}+\frac{b}{(12n+5)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{b}{(12n+7)^{5}}$ $+\frac{\sqrt{3}}{2(12n+8)^{5}}+\frac{\sqrt{2}}{2(12n+9)^{5}}+\frac{1}{2(12n+10)^{5}}+\frac{a}{(12n+11)^{5}})$. Change sign and divide: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{5}}+\frac{1}{2(12n+2)^{5}}+\frac{\sqrt{2}}{2(12n+3)^{5}}+\frac{\sqrt{3}}{2(12n+4)^{5}}+\frac{b}{(12n+5)^{5}}$ $+\frac{1}{(12n+6)^{5}}+\frac{b}{(12n+7)^{5}}+\frac{\sqrt{3}}{2(12n+8)^{5}}+\frac{\sqrt{2}}{2(12n+9)^{5}}+\frac{1}{2(12n+10)^{5}}+\frac{a}{(12n+11)^{5}})=\frac{10879}{11943936}\pi^{5}$. From (4) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}+\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}+\frac{1}{(6n+4)^{5}}+\frac{1}{(6n+5)^{5}})=(\frac{1225}{373248}+\frac{17\sqrt{3}}{279936})\pi^{5}$. Divide by $2^{5}$ and subtract: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{5}}+\frac{\sqrt{2}}{2(12n+3)^{5}}+\frac{b}{(12n+5)^{5}}+\frac{b}{(12n+7)^{5}}+\frac{\sqrt{2}}{2(12n+9)^{5}}+\frac{a}{(12n+11)^{5}})=\frac{1705}{1990656}\pi^{5}$.

Now split the resultant expression: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{5}}+\frac{\sqrt{2}}{2(12n+3)^{5}}+\frac{b}{(12n+5)^{5}}+\frac{b}{(12n+7)^{5}}+\frac{\sqrt{2}}{2(12n+9)^{5}}+\frac{a}{(12n+11)^{5}})$$=\frac{\sqrt{6}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+11)^{5}})$$-\frac{\sqrt{2}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+3)^{5}}-\frac{1}{(12n+5)^{5}}-\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+9)^{5}}+\frac{1}{(12n+11)^{5}})$$+ \frac{3\sqrt{2}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+3)^{5}}+\frac{1}{(12n+9)^{5}})$.

Let $T_{1}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+11)^{5}})$,  $T_{2}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+3)^{5}}-\frac{1}{(12n+5)^{5}}-\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+9)^{5}}+\frac{1}{(12n+11)^{5}})$ and $T_{3}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+3)^{5}}+\frac{1}{(12n+9)^{5}})$. This means that $\frac{\sqrt{6}}{4}T_{1}-\frac{\sqrt{2}}{4}T_{2}+\frac{3\sqrt{2}}{4}T_{3}=\frac{1705}{1990656}\pi^{5}$.

From (3) above we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}+\frac{1}{(4n+3)^{5}})=\frac{19\sqrt{2}}{8192}\pi^{5}$. If we divide by $3^{5}$ we get $T_{3}=\frac{19\sqrt{2}}{1990656}$.  If we instead expand the formula, we get $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}+\frac{1}{(4n+3)^{5}})=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+3)^{5}}-\frac{1}{(12n+5)^{5}}-\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+9)^{5}}+\frac{1}{(12n+11)^{5}})$, which is equal to $T_{2}$. Since $\frac{\sqrt{6}}{4}T_{1}=\frac{1705}{1990656}\pi^{5}+\frac{\sqrt{2}}{4}(T_{2}-3T_{3})=\frac{1705}{1990656}\pi^{5}+\frac{\sqrt{2}}{4}(\frac{19\sqrt{2}}{8192}-\frac{19\cdot 3\sqrt{2}}{199056})\pi^{5}=\frac{1705}{1990656}\pi^{5}=\frac{3985}{1990656}\pi^{5}$, we get $T_{1}=\frac{4}{\sqrt{6}}\cdot \frac{3985}{1990656}\pi^{5}=\frac{3985\sqrt{6}}{2985984}\pi^{5}$. Add $T_{3}:\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+3)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+9)^{5}}+\frac{1}{(12n+11)^{5}})=\frac{3985\sqrt{6}}{2985984}\pi^{5}+\frac{19\sqrt{2}}{1990656}\pi^{5}$. Now the sum $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}+\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}+\frac{1}{(6n+4)^{5}}+\frac{1}{(6n+5)^{5}})=(\frac{1225}{373248}+\frac{17\sqrt{3}}{279936})\pi^{5}$, which means that $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+2)^{5}}+\frac{1}{(12n+4)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{1}{(12n+8)^{5}}+\frac{1}{(12n+10)^{5}})=(\frac{1225}{11943936}+\frac{17\sqrt{3}}{8957952})\pi^{5}$. Add it in: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+2)^{5}}+\frac{1}{(12n+3)^{5}}+\frac{1}{(12n+4)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+8)^{5}}$$+\frac{1}{(12n+9)^{5}}+\frac{1}{(12n+10)^{5}}+\frac{1}{(12n+11)^{5}})=(\frac{1225}{11943936}+\frac{19\sqrt{2}}{1990656}+\frac{17\sqrt{3}}{8957952}+\frac{3985\sqrt{6}}{2985984})\pi^{5}$.

##### Seventh powers

$h_{7}(\frac{11\pi}{12})=\frac{31465357}{35831808}\pi^{7}=10080\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{7}}+\frac{1}{2(12n+2)^{7}}+\frac{\sqrt{2}}{2(12n+3)^{7}}+\frac{\sqrt{3}}{2(12n+4)^{7}}+\frac{b}{(12n+5)^{7}}+\frac{1}{(12n+6)^{7}}+\frac{b}{(12n+7)^{7}}$ $+\frac{\sqrt{3}}{2(12n+8)^{7}}+\frac{\sqrt{2}}{2(12n+9)^{7}}+\frac{1}{2(12n+10)^{7}}+\frac{a}{(12n+11)^{7}})$, which means that $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{7}}+\frac{1}{2(12n+2)^{7}}+\frac{\sqrt{2}}{2(12n+3)^{7}}+\frac{\sqrt{3}}{2(12n+4)^{7}}+\frac{b}{(12n+5)^{7}}+\frac{1}{(12n+6)^{7}}+\frac{b}{(12n+7)^{7}}+\frac{\sqrt{3}}{2(12n+8)^{7}}$$+\frac{\sqrt{2}}{2(12n+9)^{7}}+\frac{1}{2(12n+10)^{7}}+\frac{a}{(12n+11)^{7}})=\frac{31465357}{35831808}\frac{\pi^{7}}{10080}$. From (4) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{7}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{7}}+\frac{1}{2}\frac{1}{(6n+5)^{7}})=\frac{473935}{279936\cdot 10080}\pi^{7}$ . Divide by $2^{7}$ and subtract: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{7}}+\frac{\sqrt{2}}{2(12n+3)^{7}}+\frac{b}{(12n+5)^{7}}+\frac{b}{(12n+7)^{7}}+\frac{\sqrt{2}}{2(12n+9)^{7}}+\frac{a}{(12n+11)^{7}})$ $=(\frac{31465357}{35831808}-\frac{473935}{279936\cdot 128})\frac{\pi^{7}}{10080}=\frac{5165237}{5971968}\frac{\pi^{7}}{10080}$.

Now split the resultant expression: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{7}}+\frac{\sqrt{2}}{2(12n+3)^{7}}+\frac{b}{(12n+5)^{7}}+\frac{b}{(12n+7)^{7}}+\frac{\sqrt{2}}{2(12n+9)^{7}}+\frac{a}{(12n+11)^{7}})$$=\frac{\sqrt{6}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+5)^{7}}+\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+11)^{7}})$$-\frac{\sqrt{2}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+3)^{7}}-\frac{1}{(12n+5)^{7}}-\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+9)^{7}}+\frac{1}{(12n+11)^{7}})$$+ \frac{3\sqrt{2}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+3)^{7}}+\frac{1}{(12n+9)^{7}})$. Let $U_{1}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+5)^{7}}+\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+11)^{7}})$, let $U_{2}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+3)^{7}}-\frac{1}{(12n+5)^{7}}-\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+9)^{7}}+\frac{1}{(12n+11)^{7}})$ and let $U_{3}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+3)^{7}}+\frac{1}{(12n+9)^{7}})$. This means that $\frac{\sqrt{6}}{4}U_{1}-\frac{\sqrt{2}}{4}U_{2}+\frac{3\sqrt{2}}{4}U_{3}=\frac{5165237}{5971968}\frac{\pi^{7}}{10080}$.

From (3) above we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}+\frac{1}{(4n+3)^{7}})=\frac{19341\sqrt{2}}{8192}\frac{\pi^{7}}{10080}$. If we divide by $3^{7}$ we get $U_{3}=\frac{19341\sqrt{2}}{8192\cdot 2187}\frac{\pi^{7}}{10080}$. If we instead expand the formula, we get $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}+\frac{1}{(4n+3)^{7}})=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+3)^{7}}-\frac{1}{(12n+5)^{7}}-\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+9)^{7}}+\frac{1}{(12n+11)^{7}})$, which is equal to $U_{2}$. Since $\frac{\sqrt{6}}{4}U_{1}=\frac{5165237}{5971968}\frac{\pi^{7}}{10080}+\frac{\sqrt{2}}{4}(U_{2}-3U_{3})=\frac{5165237}{5971968}\frac{\pi^{7}}{10080}+\frac{\sqrt{2}}{4}(\frac{19341\sqrt{2}}{8192}-\frac{3\cdot 19341\sqrt{2}}{8192\cdot 2187})\frac{\pi^{7}}{10080}=\frac{12205361}{5971968}\frac{\pi^{7}}{10080}$ or $\sqrt{6}U_{1}=\frac{12205361\cdot 4}{5971968}\frac{\pi^{7}}{10080}$ giving $U_{1}=\frac{12205361\sqrt{6}}{8957952}\frac{\pi^{7}}{10080}$. Add in $U_{3}:$ $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+3)^{7}}+\frac{1}{(12n+5)^{7}}+\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+9)^{7}}+\frac{1}{(12n+11)^{7}})=\frac{12205361\sqrt{6}}{8957952}\frac{\pi^{7}}{10080}+\frac{19341\sqrt{2}}{17915904}\frac{\pi^{7}}{10080}$.

From (4) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{7}}+\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}+\frac{1}{(6n+4)^{7}}+\frac{1}{(6n+5)^{7}})=(\frac{133529}{403107840}+\frac{91\sqrt{3}}{60466176})\pi^{7}$. Divide by $2^{7}$: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+2)^{7}}+\frac{1}{(12n+4)^{7}}+\frac{1}{(12n+6)^{7}}+\frac{1}{(12n+8)^{7}}+\frac{1}{(12n+10)^{7}})=\frac{1}{128}(\frac{133529}{403107840}+\frac{91\sqrt{3}}{60466176})\pi^{7}$ and add: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+2)^{7}}+\frac{1}{(12n+3)^{7}}+\frac{1}{(12n+4)^{7}}+\frac{1}{(12n+5)^{7}}+\frac{1}{(12n+6)^{7}}+\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+8)^{7}}$$+\frac{1}{(12n+9)^{7}}+\frac{1}{(12n+10)^{7}}+\frac{1}{(12n+11)^{7}})=(\frac{12205361\sqrt{6}}{8957952}\frac{\pi^{7}}{10080}+\frac{19341\sqrt{2}}{17915904}\frac{\pi^{7}}{10080}+\frac{1}{128}(\frac{133529}{403107840}+\frac{91\sqrt{3}}{60466176})\pi^{7})$ $=(\frac{133529}{51597803520}+\frac{307\sqrt{2}}{2866544640}+\frac{91\sqrt{3}}{7739670528}+\frac{1743623\sqrt{6}}{12899450880})\pi^{7}$.

##### Ninth powers

$h_{9}(\frac{11\pi}{12})=\frac{-32641778411}{5159780352}\pi^{9}=-725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{9}}+\frac{1}{2(12n+2)^{9}}+\frac{\sqrt{2}}{2(12n+3)^{9}}+\frac{\sqrt{3}}{2(12n+4)^{9}}+\frac{b}{(12n+5)^{9}}+\frac{1}{(12n+6)^{9}}+\frac{b}{(12n+7)^{9}}$ $+\frac{\sqrt{3}}{2(12n+8)^{9}}+\frac{\sqrt{2}}{2(12n+9)^{9}}+\frac{1}{2(12n+10)^{9}}+\frac{a}{(12n+11)^{9}})$. Change sign and divide: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{9}}+\frac{1}{2(12n+2)^{9}}+\frac{\sqrt{2}}{2(12n+3)^{9}}+\frac{\sqrt{3}}{2(12n+4)^{9}}+\frac{b}{(12n+5)^{9}}$ $+\frac{1}{(12n+6)^{9}}+\frac{b}{(12n+7)^{9}}+\frac{\sqrt{3}}{2(12n+8)^{9}}+\frac{\sqrt{2}}{2(12n+9)^{9}}+\frac{1}{2(12n+10)^{9}}+\frac{a}{(12n+11)^{9}})=\frac{32641778411}{3744762188267520}\pi^{9}$. From (4) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{9}}+\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}+\frac{1}{(6n+4)^{9}}+\frac{1}{(6n+5)^{9}})=\frac{123108821}{7313988648960}\pi^{9}$. Divide by $2^{9}$ and subtract: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{9}}+\frac{\sqrt{2}}{2(12n+3)^{9}}+\frac{b}{(12n+5)^{9}}+\frac{b}{(12n+7)^{9}}+\frac{\sqrt{2}}{2(12n+9)^{9}}+\frac{a}{(12n+11)^{9}})=\frac{120439517}{13869489586176}\pi^{9} [$.

Now split the resultant expression: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{a}{(12n+1)^{9}}+\frac{\sqrt{2}}{2(12n+3)^{9}}+\frac{b}{(12n+5)^{9}}+\frac{b}{(12n+7)^{9}}+\frac{\sqrt{2}}{2(12n+9)^{9}}+\frac{a}{(12n+11)^{9}})$$=\frac{\sqrt{6}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+5)^{9}}+\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+11)^{9}})$$-\frac{\sqrt{2}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+3)^{9}}-\frac{1}{(12n+5)^{9}}-\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+9)^{9}}+\frac{1}{(12n+11)^{9}})$$+ \frac{3\sqrt{2}}{4}\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+3)^{9}}+\frac{1}{(12n+9)^{9}})$. Let $V_{1}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+5)^{9}}+\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+11)^{9}})$, $V_{2}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+3)^{9}}-\frac{1}{(12n+5)^{9}}-\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+9)^{9}}+\frac{1}{(12n+11)^{9}})$ and $V_{3}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+3)^{9}}+\frac{1}{(12n+9)^{9}})$. This means that  $\frac{\sqrt{6}}{4}V_{1}-\frac{\sqrt{2}}{4}V_{2}+\frac{3\sqrt{2}}{4}V_{3}=\frac{120439517}{13869489586176}\pi^{9}$. From (3) above we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}+\frac{1}{(4n+3)^{9}})=\frac{2256633\sqrt{2}}{95126814720}\pi^{9}$. If we divide by $3^{9}$  we get $V_{3}=\frac{83579\sqrt{2}}{69347447930880}\pi^{9}$. If we instead expand the formula, we get $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}+\frac{1}{(4n+3)^{9}})=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+3)^{9}}-\frac{1}{(12n+5)^{9}}-\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+9)^{9}}+\frac{1}{(12n+11)^{9}})$, which is equal to $V_{2}$. Since $\frac{\sqrt{6}}{4}V_{1}=\frac{120439517}{13869489586176}\pi^{9}+\frac{\sqrt{2}}{4}(V_{2}-3V_{3})=\frac{120439517}{13869489586176}\pi^{9}+\frac{\sqrt{2}}{4}(\frac{2256633\sqrt{2}}{95126814720}-\frac{3\cdot 83579\sqrt{2}}{69347447930880})\pi^{9}=\frac{202681253}{6934744793088}\pi^{9}$ , giving $V_{1}=\frac{202681253\sqrt{6}}{10402117189632}\pi^{9}$. Add in $V_{3}:$ $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+3)^{9}}+\frac{1}{(12n+5)^{9}}+\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+9)^{9}}+\frac{1}{(12n+11)^{9}})=\frac{202681253\sqrt{6}}{10402117189632}\pi^{9}+\frac{83579\sqrt{2}}{69347447930880}\pi^{9}$ .

From (4) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{9}}+\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}+\frac{1}{(6n+4)^{9}}+\frac{1}{(6n+5)^{9}})=\frac{1090549}{32506616218}\pi^{9}+\frac{207913\sqrt{3}}{5485491486720}\pi^{9}$. Divide by $2^{9}$: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+2)^{9}}+\frac{1}{(12n+4)^{9}}+\frac{1}{(12n+6)^{9}}+\frac{1}{(12n+8)^{9}}+\frac{1}{(12n+10)^{9}})=\frac{1}{512}(\frac{1090549}{32506616218}+\frac{207913\sqrt{3}}{5485491486720})\pi^{9}$and add: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+2)^{9}}+\frac{1}{(12n+3)^{9}}+\frac{1}{(12n+4)^{9}}+\frac{1}{(12n+5)^{9}}+\frac{1}{(12n+6)^{9}}+\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+8)^{9}}$$+\frac{1}{(12n+9)^{9}}+\frac{1}{(12n+10)^{9}}+\frac{1}{(12n+11)^{9}})=(\frac{1090549}{16643387503616}+\frac{83579\sqrt{2}}{69347447930880}+\frac{207913\sqrt{3}}{15552}+\frac{202681253\sqrt{6}}{10402117189632})\pi^{9}$.

## Alternating the signs

Using some other parameters, it is possible to calculate some related sums.

### 6. Actual values for x=π/3

The values of the sums in this chapter are:

• $\sum_{n=0}^{\infty}(\frac{1}{(3n+1)^{3}}-\frac{1}{(3n+2)^{3}})=\frac{4\sqrt{3}}{243}\pi^{3}$
• $\sum_{n=0}^{\infty}(\frac{1}{(3n+1)^{5}}-\frac{1}{(3n+2)^{5}})=\frac{4\sqrt{3}}{2187}\pi^{5}$
• $\sum_{n=0}^{\infty}(\frac{1}{(3n+1)^{7}}-\frac{1}{(3n+2)^{7}})=\frac{56\sqrt{3}}{295245}\pi^{7}$
• $\sum_{n=0}^{\infty}(\frac{1}{(3n+1)^{9}}-\frac{1}{(3n+2)^{9}})=\frac{3236\sqrt{3}}{167403915}\pi^{9}$

#### Proofs

The values for $\sin(\frac{2n\pi}{3})$ are ($\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2},0$). Also $\pi-\frac{2\pi}{3}=\frac{\pi}{3}$. Thus

• $h_{3}(\frac{\pi}{3})=\frac{8}{27}\pi^{3}=12\sum_{n=0}^{\infty}(\frac{\frac{\sqrt{3}}{2}}{(3n+1)^{3}}-\frac{\frac{\sqrt{3}}{2}}{(3n+2)^{3}})$. Thus $\sum_{n=0}^{\infty}(\frac{1}{(3n+1)^{3}}-\frac{1}{(3n+2)^{3}})=\frac{4\sqrt{3}}{243}\pi^{3}$
• $h_{5}(\frac{\pi}{3})=\frac{-160}{243}\pi^{5}=-240\sum_{n=0}^{\infty}(\frac{\frac{\sqrt{3}}{2}}{(3n+1)^{5}}-\frac{\frac{\sqrt{3}}{2}}{(3n+2)^{5}})$. Thus $\sum_{n=0}^{\infty}(\frac{1}{(3n+1)^{5}}-\frac{1}{(3n+2)^{5}})=\frac{4\sqrt{3}}{2187}\pi^{5}$
• $h_{7}(\frac{\pi}{3})=\frac{6272}{2187}\pi^{7}=10080\sum_{n=0}^{\infty}(\frac{\frac{\sqrt{3}}{2}}{(3n+1)^{7}}-\frac{\frac{\sqrt{3}}{2}}{(3n+2)^{7}})$. Thus $\sum_{n=0}^{\infty}(\frac{1}{(3n+1)^{7}}-\frac{1}{(3n+2)^{7}})=\frac{56\sqrt{3}}{295245}\pi^{7}$
• $h_{9}(\frac{\pi}{3})=\frac{-414208}{19683}\pi^{9}=-725760\sum_{n=0}^{\infty}(\frac{\frac{\sqrt{3}}{2}}{(3n+1)^{9}}-\frac{\frac{\sqrt{3}}{2}}{(3n+2)^{9}})$. Thus $\sum_{n=0}^{\infty}(\frac{1}{(3n+1)^{9}}-\frac{1}{(3n+2)^{9}})=\frac{3236\sqrt{3}}{167403915}\pi^{9}$

### 7. Actual values for x=π/4

The values of the sums in this chapter are:

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}-\frac{1}{(4n+2)^{3}}+\frac{1}{(4n+3)^{3}})=\frac{6\sqrt{2}-1}{256}\pi^{3}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}-\frac{1}{(4n+2)^{5}}+\frac{1}{(4n+3)^{5}})=\frac{114\sqrt{2}-5}{49152}\pi^{5}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}-\frac{1}{(4n+2)^{7}}+\frac{1}{(4n+3)^{7}})=\frac{38682\sqrt{2}-427}{165150720}\pi^{7}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}-\frac{1}{(4n+2)^{9}}+\frac{1}{(4n+3)^{9}})=(\frac{83579\sqrt{2}}{3523215360}-\frac{277}{4227858432})\pi^{9}$

#### Proofs

The values for $\sin(\frac{3n\pi}{4})$ are ($\frac{\sqrt{2}}{2}, -1, \frac{\sqrt{2}}{2}, 0, -\frac{\sqrt{2}}{2}, 1, -\frac{\sqrt{2}}{2}, 0$). Also $\pi-\frac{\pi}{4}=\frac{3\pi}{4}$.

##### Third powers

$h_{3}(\frac{\pi}{4})=\frac{15}{64}\pi^{3}=12\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{3}}-\frac{1}{(4n+2)^{3}}+\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{3}})$.

A partial result from (3) is $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}+\frac{1}{(4n+3)^{3}})=\frac{18\sqrt{2}}{12 \cdot 64}$. From (1) we know that $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{3}}=\frac{\pi^{3}}{32}$. The element we need sits in the second position and is negative, so we subtract $\frac{\pi^{3}}{32\cdot 2^{3}}=\frac{\pi^{3}}{256}$. This results in $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}-\frac{1}{(4n+2)^{3}}+\frac{1}{(4n+3)^{3}})=\frac{6\sqrt{2}-1}{256}\pi^{3}$.

##### Fifth powers

$h_{5}(\frac{\pi}{4})=\frac{-545}{1024}\pi^{5}=-240\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{5}}-\frac{1}{(4n+2)^{5}}+\frac{\frac{\sqrt{5}}{2}}{(4n+1)^{5}})$.

A partial result from (3) is $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}+\frac{1}{(4n+3)^{5}})=\frac{19\sqrt{2}}{240\cdot 8192}$. From (1) we know that $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{5}}=\frac{5\pi^{3}}{1536}$. The element we need sits in the second position and is negative, so we subtract $\frac{5\pi^{5}}{1536\cdot 2^{5}}=\frac{5\pi^{5}}{49152}$. This results in $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}-\frac{1}{(4n+2)^{5}}+\frac{1}{(4n+3)^{5}})=\frac{114\sqrt{2}-5}{49152}\pi^{5}$.

##### Seventh powers

$h_{7}(\frac{\pi}{4})=\frac{38255}{16384}\pi^{7}=10080\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{7}}-\frac{1}{(4n+2)^{7}}+\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{7}})$.

A partial result from (3) is $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}+\frac{1}{(4n+3)^{7}})=\frac{19341\sqrt{2}}{8192 \cdot 10080}$. From (1) we know that $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{7}}=\frac{61\pi^{7}}{184320}$. The element we need sits in the second position and is negative, so we subtract $\frac{61\pi^{7}}{184320\cdot 128}$. This results in $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}-\frac{1}{(4n+2)^{7}}+\frac{1}{(4n+3)^{7}})=\frac{38682\sqrt{2}-427}{165150720}\pi^{7}$.

##### Ninth powers

$h_{9}(\frac{\pi}{4})=\frac{-40325}{4096}\pi^{9}=-725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{\sqrt{2}}{2}}{(4n+1)^{9}}-\frac{1}{(4n+2)^{9}}+\frac{\frac{\sqrt{5}}{2}}{(4n+1)^{9}})$.

A partial result from (3) is $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}+\frac{1}{(4n+3)^{9}})=\frac{2256633\sqrt{2}}{131072\cdot 735760}$. From (1) we know that $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{9}}=\frac{277\pi^{9}}{8257536}$. The element we need sits in the second position and is negative, so we subtract $\frac{277\pi^{9}}{8257536\cdot 2^{9}}$. This results in $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}-\frac{1}{(4n+2)^{9}}+\frac{1}{(4n+3)^{9}})=(\frac{83579\sqrt{2}}{3523215360}-\frac{277}{4227858432})\pi^{9}$.

### 8. Actual values for x=π/6

The values of the sums in this chapter are:

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{3}}-\frac{1}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}-\frac{1}{(6n+4)^{3}}+\frac{1}{(6n+5)^{3}})=(\frac{29}{864}-\frac{5\sqrt{3}}{1944})\pi^{3}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}-\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}-\frac{1}{(6n+4)^{5}}+\frac{1}{(6n+5)^{5}})=(\frac{1225}{373248}-\frac{17\sqrt{3}}{279936})\pi^{5}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{7}}-\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}-\frac{1}{(6n+4)^{7}}+\frac{1}{(6n+5)^{7}})=(\frac{133529}{403107840}-\frac{91\sqrt{3}}{60466176})\pi^{7}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{9}}-\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}-\frac{1}{(6n+4)^{9}}+\frac{1}{(6n+5)^{9}})=(\frac{5452745}{162533081088}-\frac{27913\sqrt{3}}{10713850560\cdot 2^{9}})\pi^{9}$

#### Proofs

The values for $\sin(\frac{5n\pi}{6})$ are ($\frac{1}{2}, -\frac{\sqrt{3}}{2}, 1, -\frac{\sqrt{3}}{2}, \frac{1}{2}, 0, -\frac{1}{2}, \frac{\sqrt{3}}{2}, -1, \frac{\sqrt{3}}{2}, -\frac{1}{2}, 0,$). Also $\pi-\frac{5\pi}{6}=\frac{\pi}{6}$.

##### Third powers

$h_{3}(\frac{\pi}{6})=\frac{35}{216}\pi^{3}=12\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{1}{2}}{(6n+1)^{3}}-\frac{\frac{\sqrt{3}}{2}}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}-\frac{\frac{\sqrt{3}}{2}}{(6n+4)^{3}}+\frac{\frac{1}{2}}{(6n+5)^{3}})$.

From (1) we have $h_{3}(\frac{\pi}{2})=\frac{\pi^{3}}{32}$. It sits in position 3, so we subtract $12\cdot\frac{\pi^{3}}{32\cdot 3^{3}}=\frac{12\pi^{3}}{864}$: $\frac{35}{216}\pi^{3}-\frac{12\pi^{3}}{864}=\frac{4}{27}\pi^{3}$. The sum of the two elements close to the middle we already know from (2): $h_{3}(\frac{2\pi}{3}) =\frac{5\sqrt{3}}{243}\pi^{3}$. They sit in positions 2 and 4 and have the common factor $\frac{\sqrt{3}}{2}$, so we add them in: $\frac{4}{27}\pi^{3}+\frac{\sqrt{3}}{2}\cdot \frac{5\sqrt{3}}{243}\cdot \frac{12}{2^{3}}\pi^{3}=\frac{4}{27}\pi^{3}+\frac{5}{108}\pi^{3}=\frac{7}{36}\pi^{3}$. This means that $12\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{1}{2}}{(6n+1)^{3}}+\frac{\frac{1}{2}}{(6n+5)^{3}}) = \frac{7}{36}\pi^{3}$. Normalize: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{3}}+\frac{1}{(6n+5)^{3}}) = \frac{7}{216}\pi^{3}$. Combine the partial results: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{3}}+\frac{1}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}+\frac{1}{(6n+4)^{3}}+\frac{1}{(6n+5)^{3}})=(\frac{7}{216}-\frac{5\sqrt{3}}{243\cdot 2^{3}}+\frac{1}{864})\pi^{3}=(\frac{29}{864}-\frac{5\sqrt{3}}{1944})\pi^{3}$.

##### Fifth powers

$h_{5}(\frac{\pi}{6})=\frac{-2905}{7776}\pi^{5}=-240\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{1}{2}}{(6n+1)^{5}}-\frac{\frac{\sqrt{3}}{2}}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}-\frac{\frac{\sqrt{3}}{2}}{(6n+4)^{5}}+\frac{\frac{1}{2}}{(6n+5)^{5}})$. Change sign: $\frac{2905}{7776}\pi^{5}=240\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{1}{2}}{(6n+1)^{5}}+\frac{\frac{\sqrt{3}}{2}}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}+\frac{\frac{\sqrt{3}}{2}}{(6n+4)^{5}}+\frac{\frac{1}{2}}{(6n+5)^{5}})$.

From (1) we have $h_{5}(\frac{\pi}{2})=\frac{5\pi^{5}}{1536}$. It sits in position 3, so we subtract $240\cdot\frac{5\pi^{5}}{1536\cdot 3^{5}}=\frac{5\cdot 240\pi^{5}}{373248}$: $\frac{2905}{7776}\pi^{5}-\frac{1200\pi^{5}}{373248}=\frac{576}{373248}\pi^{5} =\frac{1}{648}\pi^{5}$. The sum of the two elements close to the middle we already know from (2): $h_{5}(\frac{2\pi}{3}) =\frac{17\sqrt{3}}{8748}\pi^{5}$. They sit in positions 2 and 4 and have the common factor $\frac{\sqrt{3}}{2}$, so we add them in: $\frac{1}{648}\pi^{5}-\frac{\sqrt{3}}{2}\cdot \frac{17\sqrt{3}}{8748}\cdot \frac{1}{2^{5}}\pi^{5}=\frac{305}{186624}\pi^{5}$. This means that $240\sum_{n=0}^{\infty}(-1)^{n}(\frac{\frac{1}{2}}{(6n+1)^{5}}+\frac{\frac{1}{2}}{(6n+5)^{5}}) = \frac{305}{186624}\pi^{5}$. Normalize: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}+\frac{1}{(6n+5)^{5}}) = \frac{305}{186624\cdot 60}\pi^{5}$. Add them all up: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}+\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}+\frac{1}{(6n+4)^{5}}+\frac{1}{(6n+5)^{5}})=(\frac{5}{373248}+\frac{17\sqrt{3}}{279936}+\frac{305}{93312})\pi^{5}=(\frac{1225}{373248}-\frac{17\sqrt{3}}{279936})\pi^{5}$.

##### Seventh powers

$h_{7}(\frac{\pi}{6})=\frac{461195}{279936}\pi^{7}=10080\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{7}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{7}}+\frac{1}{2}\frac{1}{(6n+5)^{7}})$, which means that $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{7}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{7}}+\frac{1}{2}\frac{1}{(6n+5)^{7}})=\frac{13177}{80621568}\pi^{7}$. From $h_{7}(\frac{\pi}{2})$ we can derive the middle element: $\frac{61\pi^{7}}{184320}\cdot \frac{1}{3^{7}}$. Subtract it: $\frac{13177}{80621568}\pi^{7}-\frac{61\pi^{7}}{403107840}=\frac{2057}{12597120}\pi^{7}$. The sum of the two elements close to the middle we already know from $h_{7}(\frac{2\pi}{3}):\frac{\sqrt{3}}{2}\frac{91\sqrt{3}}{472392\cdot 2^{7}}\pi^{7}=\frac{91}{40310784}\pi^{7}$. Add them in: $\frac{2057}{12597120}\pi^{7}+\frac{91}{40310784}\pi^{7}=\frac{66734}{40310784}\pi^{7}=\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{7}}+\frac{1}{2}\frac{1}{(6n+5)^{7}})$. Take the $\frac{1}{2}$ outside: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{7}}+\frac{1}{(6n+5)^{7}})=\frac{133468}{403107840}\pi^{7}$. Combine the partial results: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{7}}-\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}-\frac{1}{(6n+4)^{7}}+\frac{1}{(6n+5)^{7}})$ $=(\frac{61}{403107840}-\frac{91\sqrt{3}}{60466176}+\frac{133468}{403107840})\pi^{7}=(\frac{133529}{403107840}-\frac{91\sqrt{3}}{60466176})\pi^{7}$.

##### Ninth powers

$h_{9}(\frac{\pi}{6})=\frac{-122277169}{10077696}\pi^{9}=-725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{9}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{9}}+\frac{1}{2}\frac{1}{(6n+5)^{9}})$. Change sign: $\frac{122277169}{10077696}\pi^{9}=725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{9}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}+\frac{\sqrt{3}}{2}\frac{1}{(6n+4)^{9}}+\frac{1}{2}\frac{1}{(6n+5)^{9}})$. From $h_{9}(\frac{\pi}{2})$ we can derive the middle element: $\frac{277\pi^{9}}{8257536}\frac{1}{3^{9}}$. Subtract it: $\frac{122277169}{10077696}\pi^{9}-\frac{277\cdot 725760}{8257536}\frac{1}{3^{9}}\pi^{9}=\frac{955193}{78732}\pi^{9}$. The sum of the two elements close to the middle we already know from $h_{9}(\frac{2\pi}{3}): \frac{\sqrt{3}}{2}\frac{207913\sqrt{3}}{10713850560\cdot 2^{9}}\pi^{9}=\frac{207913}{7142567040\cdot 512}\pi^{9}$. Add them in: $(\frac{955193}{78732}+\frac{16840953}{408146688})\pi^{9}=\frac{6815585}{559872}\pi^{9}$. Thus $725760\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{2}\frac{1}{(6n+1)^{9}}+\frac{1}{2}\frac{1}{(6n+5)^{9}})=\frac{6815585}{559872}\pi^{9}$. Take the ½ outside: $362880\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{9}}+\frac{1}{(6n+5)^{9}})=\frac{6815585}{559872}\pi^{9}$. Combine the partial results: $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{9}}-\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}-\frac{1}{(6n+4)^{9}}+\frac{1}{(6n+5)^{9}})$ $=(\frac{277}{8257536}\cdot\frac{1}{3^{9}}-\frac{207913\sqrt{3}}{10713850560}\cdot\frac{1}{2^{9}}+\frac{6815585}{559872\cdot 362880})\pi^{9}=(\frac{5452745}{162533081088}-\frac{27913\sqrt{3}}{10713850560\cdot 2^{9}})\pi^{9}$.

### 9. Actual values for x=π/12

The values in this chapter are:

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}-\frac{1}{(12n+2)^{3}}+\frac{1}{(12n+3)^{3}}-\frac{1}{(12n+4)^{3}}+\frac{1}{(12n+5)^{3}}- \frac{1}{(12n+6)^{3}}+\frac{1}{(12n+7)^{3}}$$-\frac{1}{(12n+8)^{3}}+\frac{1}{(12n+9)^{3}}-\frac{1}{(12n+10)^{3}}+\frac{1}{(12n+11)^{3}})=(-\frac{29}{6912}+\frac{\sqrt{2}}{1152}-\frac{5\sqrt{3}}{15552}+\frac{23\sqrt{6}}{1728})\pi^{3}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}-\frac{1}{(12n+2)^{5}}+\frac{1}{(12n+3)^{5}}-\frac{1}{(12n+4)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+8)^{5}}$$+\frac{1}{(12n+9)^{5}}-\frac{1}{(12n+10)^{5}}+\frac{1}{(12n+11)^{5}})=(-\frac{1225}{11943936}+\frac{19\sqrt{2}}{1990656}-\frac{17\sqrt{3}}{8957952}+\frac{3985\sqrt{6}}{2985984})\pi^{5}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}-\frac{1}{(12n+2)^{7}}+\frac{1}{(12n+3)^{7}}-\frac{1}{(12n+4)^{7}}+\frac{1}{(12n+5)^{7}}-\frac{1}{(12n+6)^{7}}+\frac{1}{(12n+7)^{7}}-\frac{1}{(12n+8)^{7}}$$+\frac{1}{(12n+9)^{7}}-\frac{1}{(12n+10)^{7}}+\frac{1}{(12n+11)^{7}})=(-\frac{133529}{51597803520}+\frac{307\sqrt{2}}{2866544640}-\frac{91\sqrt{3}}{7739670528}+\frac{1743623\sqrt{6}}{12899450880})\pi^{7}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}-\frac{1}{(12n+2)^{9}}+\frac{1}{(12n+3)^{9}}-\frac{1}{(12n+4)^{9}}+\frac{1}{(12n+5)^{9}}-\frac{1}{(12n+6)^{9}}+\frac{1}{(12n+7)^{9}}-\frac{1}{(12n+8)^{9}}$$+\frac{1}{(12n+9)^{9}}-\frac{1}{(12n+10)^{9}}+\frac{1}{(12n+11)^{9}})=(-\frac{1090549}{16643387503616}+\frac{83579\sqrt{2}}{69347447930880}-\frac{207913\sqrt{3}}{2808571641200640}+\frac{202681253\sqrt{6}}{10402117189632})\pi^{9}$

#### Proofs

The values for $\sin(\frac{11\pi}{12})$ are ($a, -\frac{1}{2}, \frac{\sqrt{2}}{2}, -\frac{\sqrt{3}}{2}, b, -1, b, -\frac{\sqrt{3}}{2}, \frac{\sqrt{2}}{2}, -\frac{1}{2}, a, 0$), where  $a=\sin(\frac{\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4}$ and  $b=\sin(\frac{5\pi}{12})=\frac{\sqrt{6}+\sqrt{2}}{4}$. Of course  $\pi-\frac{–\pi}{12}=\frac{\pi}{12}$. Thus:

##### Third powers

From (5) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+3)^{3}}+\frac{1}{(12n+5)^{3}}+\frac{1}{(12n+7)^{3}}+\frac{1}{(12n+9)^{3}}+\frac{1}{(12n+11)^{3}})=\frac{23\sqrt{6}}{1728}\pi^{3}$ and $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+2)^{3}}+\frac{1}{(12n+4)^{3}}+\frac{1}{(12n+6)^{3}}+\frac{1}{(12n+8)^{3}}+\frac{1}{(12n+10)^{3}})=\frac{261+20\sqrt{3}}{62208}\pi^{3}$. Subtraction gives $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}-\frac{1}{(12n+2)^{3}}+\frac{1}{(12n+3)^{3}}-\frac{1}{(12n+4)^{3}}+\frac{1}{(12n+5)^{3}}- \frac{1}{(12n+6)^{3}}+\frac{1}{(12n+7)^{3}}$$-\frac{1}{(12n+8)^{3}}+\frac{1}{(12n+9)^{3}}-\frac{1}{(12n+10)^{3}}+\frac{1}{(12n+11)^{3}})=(-\frac{29}{6912}+\frac{\sqrt{2}}{1152}-\frac{5\sqrt{3}}{15552}+\frac{23\sqrt{6}}{1728})\pi^{3}$.

##### Fifth powers

From (5) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+3)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+9)^{5}}+\frac{1}{(12n+11)^{5}})=\frac{3985\sqrt{6}}{2985984}\pi^{5}+\frac{19\sqrt{2}}{1990656}\pi^{5}$ and $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+2)^{5}}+\frac{1}{(12n+4)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{1}{(12n+8)^{5}}+\frac{1}{(12n+10)^{5}})=(\frac{1225}{11943936}+\frac{17\sqrt{3}}{8957952})\pi^{5}$. Subtraction gives $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}-\frac{1}{(12n+2)^{5}}+\frac{1}{(12n+3)^{5}}-\frac{1}{(12n+4)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+8)^{5}}$$+\frac{1}{(12n+9)^{5}}-\frac{1}{(12n+10)^{5}}+\frac{1}{(12n+11)^{5}})=(-\frac{1225}{11943936}+\frac{19\sqrt{2}}{1990656}-\frac{17\sqrt{3}}{8957952}+\frac{3985\sqrt{6}}{2985984})\pi^{5}$.

##### Seventh powers

From (5) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+3)^{7}}+\frac{1}{(12n+5)^{7}}+\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+9)^{7}}+\frac{1}{(12n+11)^{7}})=\frac{12205361\sqrt{6}}{8957952}\frac{\pi^{7}}{10080}+\frac{19341\sqrt{2}}{17915904}\frac{\pi^{7}}{10080}$ and $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+2)^{7}}+\frac{1}{(12n+4)^{7}}+\frac{1}{(12n+6)^{7}}+\frac{1}{(12n+8)^{7}}+\frac{1}{(12n+10)^{7}})=\frac{1}{128}(\frac{133529}{403107840}+\frac{91\sqrt{3}}{60466176})\pi^{7}$. Subtraction gives $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}-\frac{1}{(12n+2)^{7}}+\frac{1}{(12n+3)^{7}}-\frac{1}{(12n+4)^{7}}+\frac{1}{(12n+5)^{7}}-\frac{1}{(12n+6)^{7}}+\frac{1}{(12n+7)^{7}}-\frac{1}{(12n+8)^{7}}$ $+\frac{1}{(12n+9)^{7}}-\frac{1}{(12n+10)^{7}}+\frac{1}{(12n+11)^{7}})=(-\frac{133529}{51597803520}+\frac{19341\sqrt{2}}{17915904\cdot10080}-\frac{91\sqrt{3}}{7739670528}+\frac{12205361\sqrt{6}}{90296156160})\pi^{7}$

##### Ninth powers

From (5) we have $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+3)^{9}}+\frac{1}{(12n+5)^{9}}+\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+9)^{9}}+\frac{1}{(12n+11)^{9}})=\frac{202681253\sqrt{6}}{10402117189632}\pi^{9}+\frac{83579\sqrt{2}}{69347447930880}\pi^{9}$ and $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+2)^{9}}+\frac{1}{(12n+4)^{9}}+\frac{1}{(12n+6)^{9}}+\frac{1}{(12n+8)^{9}}+\frac{1}{(12n+10)^{9}})=\frac{1}{512}(\frac{1090549}{32506616218}+\frac{207913\sqrt{3}}{5485491486720})\pi^{9}$. Subtraction gives $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}-\frac{1}{(12n+2)^{9}}+\frac{1}{(12n+3)^{9}}-\frac{1}{(12n+4)^{9}}+\frac{1}{(12n+5)^{9}}-\frac{1}{(12n+6)^{9}}+\frac{1}{(12n+7)^{9}}-\frac{1}{(12n+8)^{9}}$$+\frac{1}{(12n+9)^{9}}-\frac{1}{(12n+10)^{9}}+\frac{1}{(12n+11)^{9}})$ $=\frac{202681253\sqrt{6}}{10402117189632}\pi^{9}+\frac{83579\sqrt{2}}{69347447930880}\pi^{9}-\frac{1}{512}(\frac{1090549}{32506616218}+\frac{207913\sqrt{3}}{5485491486720})\pi^{9}$ $=(-\frac{1090549}{16643387503616}+\frac{83579\sqrt{2}}{69347447930880}-\frac{207913\sqrt{3}}{2808571641200640}+\frac{202681253\sqrt{6}}{10402117189632})\pi^{9}$

## Filling in the zeroes

In all the series above, there are regularly spaced zeroes. It is possible to fill them in by multiplying the non-zero values by a power of the index of the first zero and add to the original series. This must be done repeatedly using increasing multiples of the first factor. These additions form a geometric series, which means that it is possible to find the value of the series. The only drawback is that the sign of the new elements does not follow those of the neighbors.

### 10. Additional values for x=π/2

The values for $\sin(\frac{n\pi}{2})$ are (1, 0, -1, 0) which means that the index of the first zero is 2. In this case the factor will be 2-p and the sum of the geometric series will be $\frac{1}{1-2^{-p}}=\frac{2^{p}}{2^{p}-1}$. This means that

$\sum_{n=0}^{\infty}(\frac{(-1)^{n}}{(2n+1)^{3}}+\frac{(-1)^{q}}{(2n+2)^{3}})=\frac{\pi^{3}}{32}\cdot\frac{8}{7}=\frac{\pi^{3}}{28}$
$\sum_{n=0}^{\infty}(\frac{(-1)^{n}}{(2n+1)^{5}}+\frac{(-1)^{q}}{(2n+2)^{5}})=\frac{5\pi^{5}}{1536}\cdot\frac{32}{31}=\frac{5\pi^{5}}{1488}$
$\sum_{n=0}^{\infty}(\frac{(-1)^{n}}{(2n+1)^{7}}+\frac{(-1)^{q}}{(2n+2)^{7}})=\frac{61\pi^{7}}{184320}\cdot\frac{128}{127}=\frac{61\pi^{7}}{182880}$
$\sum_{n=0}^{\infty}(\frac{(-1)^{n}}{(2n+1)^{9}}+\frac{(-1)^{q}}{(2n+2)^{9}})=\frac{277\pi^{9}}{8257536}\cdot\frac{512}{511}=\frac{277\pi^{7}}{8241408}$

It is possible to calculate the value of the integer q, but the answer is not very informative.

### 11. Additional values for x=2π/3

The values for $\sin(\frac{n\pi}{3})$ are ($\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}, 0, -\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}, 0$) which means that the index of the first zero is 3. In this case the factor will be 3-p and the sum of the geometric series will be $\frac{1}{1-3^{-p}}=\frac{3^{p}}{3^{p}-1}$. This means that

$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(3n+1)^{3}}+\frac{1}{(3n+2)^{3}}+\frac{(-1)^{q}}{(3n+3)^{3}})=\frac{5\sqrt{3}}{243}\cdot\frac{27}{26}\pi^{3}=\frac{5\sqrt{3}}{234}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(3n+1)^{5}}+\frac{1}{(3n+2)^{5}}+\frac{(-1)^{q}}{(3n+3)^{5}})=\frac{17\sqrt{3}}{8748}\cdot\frac{2187}{2186}\pi^{5}=\frac{17\sqrt{3}}{8744}\pi^{5}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(3n+1)^{7}}+\frac{1}{(3n+2)^{7}}+\frac{(-1)^{q}}{(3n+3)^{7}})=\frac{91\sqrt{3}}{472392}\cdot\frac{19683}{19682}\pi^{7}=\frac{91\sqrt{3}}{472368}\pi^{7}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(3n+1)^{9}}+\frac{1}{(3n+2)^{9}}+\frac{(-1)^{q}}{(3n+3)^{9}})=\frac{207913\sqrt{3}}{10713850560}\cdot\frac{177147}{177146}\pi^{9}=\frac{9207913\sqrt{3}}{10713790080}\pi^{9}$

### 12. Additional values for x=3π/4

The values for $\sin(\frac{n\pi}{4})$  are ($\frac{\sqrt{2}}{2}, 1, \frac{\sqrt{2}}{2}, 0, -\frac{\sqrt{2}}{2}, -1, -\frac{\sqrt{2}}{2}, 0$) which means that the index of the first zero is 4. In this case the factor will be 4-p and the sum of the geometric series will be $\frac{1}{1-4^{-p}}=\frac{4^{p}}{4^{p}-1}$. This means that

$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{3}}+\frac{1}{(4n+2)^{3}}+\frac{1}{(4n+3)^{3}}+\frac{(-1)^{q}}{(4n+4)^{3}})=\frac{1+6\sqrt{2}}{256}\cdot\frac{64}{63}\pi^{3}=\frac{1+6\sqrt{2}}{252}\pi^{3}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{5}}+\frac{1}{(4n+2)^{5}}+\frac{1}{(4n+3)^{5}}+\frac{(-1)^{q}}{(4n+4)^{5}})=\frac{5+114\sqrt{2}}{49152}\cdot\frac{1024}{1023}\pi^{5}=\frac{5+114\sqrt{2}}{49104}\pi^{5}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{7}}+\frac{1}{(4n+2)^{7}}+\frac{1}{(4n+3)^{7}}+\frac{(-1)^{q}}{(4n+4)^{7}})=\frac{427+38682\sqrt{2}}{165150720}\cdot\frac{16384}{16383}\pi^{7}=\frac{427+38682\sqrt{2}}{165140640}\pi^{7}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(4n+1)^{9}}+\frac{1}{(4n+2)^{9}}+\frac{1}{(4n+3)^{9}}+\frac{(-1)^{q}}{(4n+4)^{9}})=(\frac{277}{4227858432}+\frac{83579\sqrt{2}}{3523215360})\cdot\frac{262144}{262143}\pi^{9}=(\frac{277}{4227842304}+\frac{83579\sqrt{2}}{3523201920})\pi^{9}$

### 13. Additional values for x=5π/6

The values for $\sin(\frac{n\pi}{6})$ are ($\frac{1}{2}, \frac{\sqrt{3}}{2}, 1, \frac{\sqrt{3}}{2}, \frac{1}{2}, 0, -\frac{1}{2}, -\frac{\sqrt{3}}{2}, -1, -\frac{\sqrt{3}}{2}, -\frac{1}{2}, 0,$) which means that the index of the first zero is 6. In this case the factor will be 6-p and the sum of the geometric series will be $\frac{1}{1-6^{-p}}=\frac{6^{p}}{6^{p}-1}$. This means that

$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{6n+1)^{3}}+\frac{1}{6n+2)^{3}}+\frac{1}{6n+3)^{3}}+\frac{1}{6n+4)^{3}}+\frac{1}{6n+5)^{3}}+\frac{(-1)^{q}}{6n+6)^{3}})=\frac{261+20\sqrt{3}}{7776}\cdot\frac{216}{215}\pi^{3}=\frac{261+20\sqrt{3}}{7740}\pi^{3}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{6n+1)^{5}}+\frac{1}{6n+2)^{5}}+\frac{1}{6n+3)^{5}}+\frac{1}{6n+4)^{5}}+\frac{1}{6n+5)^{5}}+\frac{(-1)^{q}}{6n+6)^{5}})=(\frac{1225}{373248}+\frac{17\sqrt{3}}{279936})\cdot\frac{7776}{7775}\pi^{5}=(\frac{49}{14928}+\frac{17\sqrt{3}}{279900})\pi^{5}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{6n+1)^{7}}+\frac{1}{6n+2)^{7}}+\frac{1}{6n+3)^{7}}+\frac{1}{6n+4)^{7}}+\frac{1}{6n+5)^{7}}+\frac{(-1)^{q}}{6n+6)^{7}})=(\frac{133529}{403107840}+\frac{91\sqrt{3}}{60466176})\cdot\frac{279936}{279935}\pi^{7}$ $=(\frac{133529}{403106400}+\frac{91\sqrt{3}}{60465960})\pi^{7}$
$\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{6n+1)^{9}}+\frac{1}{6n+2)^{9}}+\frac{1}{6n+3)^{9}}+\frac{1}{6n+4)^{9}}+\frac{1}{6n+5)^{9}}+\frac{(-1)^{q}}{6n+6)^{9}})$ $=(\frac{1090549}{32506616218}+\frac{207913\sqrt{3}}{10713850560\cdot 512})\cdot\frac{10077696}{10077695}\pi^{9}$

### 14. Additional values for x=11π/12

The index of the first zero is 12. In this case the factor will be $12^{-p}$ and the sum of the geometric series will be $\frac{1}{1-12^{-p}}=\frac{12^{p}}{12^{p}-1}$. This means that:

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}+\frac{1}{(12n+2)^{3}}+\frac{1}{(12n+3)^{3}}+\frac{1}{(12n+4)^{3}}+\frac{1}{(12n+5)^{3}}+ \frac{1}{(12n+6)^{3}}+\frac{1}{(12n+7)^{3}}$$+\frac{1}{(12n+8)^{3}}+\frac{1}{(12n+9)^{3}}+\frac{1}{(12n+10)^{3}}+\frac{1}{(12n+11)^{3}}+\frac{(-1)^{q}}{(12n+12)^{3}})=(\frac{29}{6912}+\frac{\sqrt{2}}{1152}+\frac{5\sqrt{3}}{15552}+\frac{23\sqrt{6}}{1728})\pi^{3}\cdot \frac{1728}{1727}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}+\frac{1}{(12n+2)^{5}}+\frac{1}{(12n+3)^{5}}+\frac{1}{(12n+4)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+8)^{5}}$$+\frac{1}{(12n+9)^{5}}+\frac{1}{(12n+10)^{5}}+\frac{1}{(12n+11)^{5}}+\frac{(-1)^{q}}{(12n+12)^{5}})=(\frac{1225}{11943936}+\frac{19\sqrt{2}}{1990656}+\frac{17\sqrt{3}}{8957952}+\frac{3985\sqrt{6}}{2985984})\pi^{5}\cdot\frac{248832}{248831}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}+\frac{1}{(12n+2)^{7}}+\frac{1}{(12n+3)^{7}}+\frac{1}{(12n+4)^{7}}+\frac{1}{(12n+5)^{7}}+\frac{1}{(12n+6)^{7}}+\frac{1}{(12n+7)^{7}}+\frac{1}{(12n+8)^{7}}$$+\frac{1}{(12n+9)^{7}}+\frac{1}{(12n+10)^{7}}+\frac{1}{(12n+11)^{7}}+\frac{(-1)^{q}}{(12n+12)^{7}})=(\frac{133529}{51597803520}+\frac{307\sqrt{2}}{2866544640}+\frac{91\sqrt{3}}{7739670528}+\frac{1743623\sqrt{6}}{12899450880})\pi^{7}\cdot\frac{35831808}{35831807}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}+\frac{1}{(12n+2)^{9}}+\frac{1}{(12n+3)^{9}}+\frac{1}{(12n+4)^{9}}+\frac{1}{(12n+5)^{9}}+\frac{1}{(12n+6)^{9}}+\frac{1}{(12n+7)^{9}}+\frac{1}{(12n+8)^{9}}$$+\frac{1}{(12n+9)^{9}}+\frac{1}{(12n+10)^{9}}+\frac{1}{(12n+11)^{9}}+\frac{(-1)^{q}}{(12n+12)^{9}})$ $=(\frac{1090549}{16643387503616}+\frac{83579\sqrt{2}}{69347447930880}+\frac{207913\sqrt{3}}{15552}+\frac{202681253\sqrt{6}}{10402117189632})\pi^{9}\cdot\frac{5159780352}{5159780351}$

## Approximating the Dirichlet eta function

In the cases where the first zero occurs at an even number, we can subtract the positive geometric series instead of adding it in order to get slightly closer to the eta function values.

#### Optimized version for x=6

The correction factor here is $\frac{1}{6^{p}-1}$

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{3}}-\frac{1}{(6n+2)^{3}}+\frac{1}{(6n+3)^{3}}-\frac{1}{(6n+4)^{3}}+\frac{1}{(6n+5)^{3}}-\frac{(-1)^{q}}{(6n+6)^{3}})$ $=(\frac{29}{864}-\frac{5\sqrt{3}}{1944})\pi^{3}-\frac{1}{215}\frac{261+20\sqrt{3}}{7776}\pi^{3}=(\frac{2059}{61920}-\frac{430\sqrt{3}}{167184})\pi^{3}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{5}}-\frac{1}{(6n+2)^{5}}+\frac{1}{(6n+3)^{5}}-\frac{1}{(6n+4)^{5}}+\frac{1}{(6n+5)^{5}}-\frac{(-1)^{q}}{(6n+6)^{5}})$ $=(\frac{1225}{373248}-\frac{17\sqrt{3}}{279936})\pi^{5}-\frac{1}{7775}(\frac{1225}{373248}+\frac{17\sqrt{3}}{279936})\pi^{5}=(\frac{190463}{58040064}-\frac{66079\sqrt{3}}{1088251200})\pi^{5}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{7}}-\frac{1}{(6n+2)^{7}}+\frac{1}{(6n+3)^{7}}-\frac{1}{(6n+4)^{7}}+\frac{1}{(6n+5)^{7}}-\frac{(-1)^{q}}{(6n+6)^{7}})$ $=(\frac{133529}{403107840}-\frac{91\sqrt{3}}{60466176})\pi^{7}-\frac{1}{279935}(\frac{133529}{403107840}+\frac{91\sqrt{3}}{60466176})\pi^{7}=(\frac{18689653543}{56421996595200}-\frac{91\sqrt{3}}{60465960})\pi^{7}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(6n+1)^{9}}-\frac{1}{(6n+2)^{9}}+\frac{1}{(6n+3)^{9}}-\frac{1}{(6n+4)^{9}}+\frac{1}{(6n+5)^{9}}-\frac{(-1)^{q}}{(6n+6)^{9}})$ $=(\frac{5452745}{162533081088}-\frac{27913\sqrt{3}}{10713850560\cdot 2^{9}})\pi^{9}-\frac{1}{10077695}(\frac{1090549}{32506616218}+\frac{207913\sqrt{3}}{5485491486720)})\pi^{9}$ $=(\frac{893137087226802577463819}{26622249348805387686234685440}-\frac{207913\sqrt{3}}{5485490942400})\pi^{9}$
##### Some error estimates

The main error in the sums above lies in the terms from 7 to 12 which have the wrong sign. The errors introduced here are:

p=3: The main error is 0.00301336.

p=5: The main error is 7.62139E-05

p=7: The main error is 1.73982E-06

p=9: The main error is 3.82836E-08

#### Optimized version for x=12

The correction factor here is $\frac{1}{6^{p}-1}$.

• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{3}}-\frac{1}{(12n+2)^{3}}+\frac{1}{(12n+3)^{3}}-\frac{1}{(12n+4)^{3}}+\frac{1}{(12n+5)^{3}}- \frac{1}{(12n+6)^{3}}+\frac{1}{(12n+7)^{3}}$$-\frac{1}{(12n+8)^{3}}+\frac{1}{(12n+9)^{3}}-\frac{1}{(12n+10)^{3}}+\frac{1}{(12n+11)^{3}}-\frac{(-1)^{q}}{(12n+12)^{3}})$ $=(-\frac{29}{6912}+\frac{\sqrt{2}}{1152}-\frac{5\sqrt{3}}{15552}+\frac{23\sqrt{6}}{1728})\pi^{3}-\frac{1}{1727}(\frac{29}{6912}+\frac{\sqrt{2}}{1152}+\frac{5\sqrt{3}}{15552}+\frac{23\sqrt{6}}{1728})\pi^{3}$ $=(-\frac{29}{6908}+\frac{863\sqrt{2}}{994752}-\frac{5\sqrt{3}}{15543}+\frac{19849\sqrt{6}}{1492128}\pi^{3}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{5}}-\frac{1}{(12n+2)^{5}}+\frac{1}{(12n+3)^{5}}-\frac{1}{(12n+4)^{5}}+\frac{1}{(12n+5)^{5}}+\frac{1}{(12n+6)^{5}}+\frac{1}{(12n+7)^{5}}+\frac{1}{(12n+8)^{5}}$$+\frac{1}{(12n+9)^{5}}-\frac{1}{(12n+10)^{5}}+\frac{1}{(12n+11)^{5}}-\frac{(-1)^{q}}{(12n+12)^{5}})=$ $(-\frac{1225}{11943936}+\frac{19\sqrt{2}}{1990656}-\frac{17\sqrt{3}}{8957952}+\frac{3985\sqrt{6}}{2985984})\pi^{5}$ $-\frac{1}{248831}(\frac{1225}{11943936}+\frac{19\sqrt{2}}{1990656}+\frac{17\sqrt{3}}{8957952}+\frac{3985\sqrt{6}}{2985984})\pi^{5}=(-\frac{1225}{11943888}+\frac{2363885\sqrt{2}}{247668461568}-\frac{17\sqrt{3}}{8957916}+\frac{495793775\sqrt{6}}{371502692352})\pi^{5}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{7}}-\frac{1}{(12n+2)^{7}}+\frac{1}{(12n+3)^{7}}-\frac{1}{(12n+4)^{7}}+\frac{1}{(12n+5)^{7}}-\frac{1}{(12n+6)^{7}}+\frac{1}{(12n+7)^{7}}-\frac{1}{(12n+8)^{7}}$$+\frac{1}{(12n+9)^{7}}-\frac{1}{(12n+10)^{7}}+\frac{1}{(12n+11)^{7}}-\frac{(-1)^{q}}{(12n+12)^{7}})$ $=(-\frac{133529}{51597803520}+\frac{307\sqrt{2}}{2866544640}-\frac{91\sqrt{3}}{7739670528}+\frac{1743623\sqrt{6}}{12899450880})\pi^{7}$ $-\frac{1}{35831807}(\frac{133529}{51597803520}+\frac{307\sqrt{2}}{2866544640}+\frac{91\sqrt{3}}{7739670528}+\frac{1743623\sqrt{6}}{12899450880})\pi^{7}$ $=(-\frac{12139}{4690709280}+\frac{307\sqrt{2}}{2866544560}-\frac{1630347173\sqrt{3}}{138663190301442048}+\frac{31238580536569\sqrt{6}}{231105317169070080})\pi^{7}$
• $\sum_{n=0}^{\infty}(-1)^{n}(\frac{1}{(12n+1)^{9}}-\frac{1}{(12n+2)^{9}}+\frac{1}{(12n+3)^{9}}-\frac{1}{(12n+4)^{9}}+\frac{1}{(12n+5)^{9}}-\frac{1}{(12n+6)^{9}}+\frac{1}{(12n+7)^{9}}-\frac{1}{(12n+8)^{9}}$$+\frac{1}{(12n+9)^{9}}-\frac{1}{(12n+10)^{9}}+\frac{1}{(12n+11)^{9}}-\frac{(-1)^{q}}{(12n+12)^{9}})$ $=(-\frac{1090549}{16643387503616}+\frac{83579\sqrt{2}}{69347447930880}-\frac{207913\sqrt{3}}{2808571641200640}+\frac{202681253\sqrt{6}}{10402117189632})\pi^{9}$ $-\frac{1}{5159780351}(\frac{1090549}{16643387503616}+\frac{83579\sqrt{2}}{69347447930880}+\frac{207913\sqrt{3}}{15552}+\frac{202681253\sqrt{6}}{10402117189632})\pi^{9}$ $=(-\frac{5495110647552}{83863499819567166259}+\frac{43124928187265\sqrt{2}}{35781759922575023013888}-\frac{207913\sqrt{3}}{2808571640656320}+\frac{522895373271389275\sqrt{6}}{26836319941931267260416})\pi^{9}$.
##### Some error estimates

The main error in the sums above lies in the terms from 13 to 23 which have the wrong sign. The errors introduced here are:

• p=3: The main error is 9.08E-04
• p=5: The main error is 3.34E-06
• p=7: The main error is 2.03E-08
• p=9: The main error is 6.29E-11

## What now?

I would like to take the next step, but I think I will have to find another method. Essentially I am calculating $\sum_{n=0}^{\infty}\sin(\frac{n}{M}\pi)\frac{1}{n^{p}}$ for increasingly large M. A clever way of calculating the next sine would help – or would it? I am going to let the problem simmer in my brain for a while.

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