How to find the sum of this series?

[Rick16]

TL;DR

I have a series and know its sum but I have no idea how to get there.

I have been working through problem 2.7 in Griffiths' QM book. Everything is pretty much clear, except that at the very end the following sum shows up: $(1/1 + 1/3^2 + 1/5^2 +1/7^2 +...)$ According to the solution, this sum equals $\pi^2/8$. How can this be determined?

 

[.Scott]

This is related to the Basel Problem. Starting from there, you can separate the odd and even terms, then note that the even terms are 1/4 of the odd terms, and work it from there.

Deriving a solution to the Basel Problem is not something that would be assigned to students. It was over 8 decades before Euler finally demonstrated the solution.

It is related to the Taylor series for sin.

 

[Rick16]

So apparently there is no straight forward algorithm for this. I will look for an online calculator then.

 

[phyzguy]

If you know the answer, why do you need an online calculator?

 

[julian]

So apparently there is no straight forward algorithm for this. I will look for an online calculator then.

In another thread I provide a derivation of the general closed formula for $\zeta(2k)$ in terms of Bernoulli numbers:

\begin{align*}
\zeta(2k) = \sum_{n=1}^\infty \frac{1}{n^{2k}} = (-1)^{k+1} (2 \pi)^{2k} \dfrac{B_{2k}}{2 (2k)!}
\end{align*}

From which you can obatin your series as explained by @pasmith in this other thread. The thread is found here: Question on an infinite summation series. This derivation, contained in the attachment, uses complex contour integration directly.

In a second method, presented in the thread itself, relates the infinite sum to a real integral, which is then evaluated using contour integration to obtain:

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^{2k}} & = (1-2^{-2k}) (-1)^{k+1} (2 \pi)^{2k} \frac{B_{2k}}{2 (2k)!} .
\end{align*}

In the second method, I use an identity that can be derived more directly than in the derivation given in the thread:

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx - \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} & = \int_{-\infty}^\infty \dfrac{4 x^{2k} e^{2x}}{(e^{2x} - 1)^2} dx
\nonumber \\
& = \int_{-\infty}^\infty \dfrac{2^{-2k+1} (2x)^{2k} e^{2x}}{(e^{2x} - 1)^2} 2dx
\nonumber \\
& = 2^{-2k+1} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx
\end{align*}

implies

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx = \frac{1}{1- 2^{-2k+1}} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2}
\end{align*}

 

[julian]

Extracting relevant bits from other thread: We can write

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \sum_{n=1}^\infty \dfrac{1}{n^2} - \sum_{n=1}^\infty \dfrac{1}{(2n)^2}
\nonumber \\
& = (1 - 2^{-2}) \sum_{n=1}^\infty \dfrac{1}{n^2}
\nonumber \\
& = \frac34 \sum_{n=1}^\infty \dfrac{1}{n^2}
\end{align*}

We have

\begin{align*}
\sum_{n=1}^\infty \dfrac{1}{n^2} & = \sum_{n=1}^\infty \dfrac{1}{n^2} \int_0^\infty e^{-y} y dy
\nonumber \\
& = \sum_{n=1}^\infty \int_0^\infty e^{-nx} x dx
\nonumber \\
& = \int_0^\infty \dfrac{x}{e^{x} - 1} dx
\nonumber \\
& = \frac{1}{2} \int_0^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx
\nonumber \\
& = \frac{1}{4} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx
\end{align*}

So

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \frac{3}{16} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx \quad (*)
\end{align*}

Note that

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx - \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} & = \int_{-\infty}^\infty \dfrac{4 x^2 e^{2x}}{(e^{2x} - 1)^2} dx
\nonumber \\
& = \int_{-\infty}^\infty \dfrac{2^{-2+1} (2x)^2 e^{2x}}{(e^{2x} - 1)^2} 2dx
\nonumber \\
& = 2^{-2+1} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx
\end{align*}

implies

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx = 2 \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2}
\end{align*}

Substituting this into $(*)$,

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \frac38 \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} dx
\end{align*}

Consider the integral:

\begin{align*}
\int_{-\infty}^\infty \dfrac{e^{\alpha x} e^x}{(e^{x} + 1)^2} dx
\end{align*}

where $-\frac{1}{2} \leq \alpha \leq \frac{1}{2}$. Then

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx = \left. \dfrac{\partial^2}{\partial \alpha^2} \int_{-\infty}^\infty \dfrac{e^{\alpha x} e^x}{(e^{x} + 1)^2} dx \right|_{\alpha=0}
\end{align*}

You can evaluate this integral using complex analysis by considering the rectangular contour, $C$, in the figure

and the integral

\begin{align*}
\oint_C \dfrac{e^{\alpha z} e^z}{(e^z + 1)^2} dz
\end{align*}

whose integrand has a pole at $\pi i$.

 

[Rick16]

If you know the answer, why do you need an online calculator?

I obviously don't need it in this case. I will need it for similar questions that may come up in the future.

 

[Rick16]

In another thread I provide a derivation of the general closed formula for ζ(2k) in terms of Bernoulli numbers:

Thank you for your detailed answer. It is, however, somewhat over my head. When I asked the question I had no idea that this was such a tricky business.

 

[PhDeezNutz]

I am also going back through Griffiths Quantum (3rd edition) after years off...........you should not get bogged down in the entire theory of Riemann Zeta functions. You should just be able to recognize them, look their values up in a table, and then mathematically move them around to solve for the series in question.

The Riemann Zeta function

$\zeta(2) = \sum\limits_{k=1}^{\infty} \frac{1}{k^2}$

The problem is that you only want the terms with odd numbers squared in the denominator

So break it into even and odd parts

$\zeta(2) = \sum\limits_{k=1}^{\infty} \frac{1}{k^2} = \sum\limits_{n=1}^{\infty} \frac{1}{(2n)^2} + \sum\limits_{n=1}^{\infty} \frac{1}{(2n-1)^2}$

$\zeta(2) = \frac{1}{4} \zeta(2) + \sum\limits_{n=1}^{\infty} \frac{1}{(2n-1)^2}$

look up the value of $\zeta(2) = \frac{pi^2}{6}$ and just solve for the series you need.


Even Griffiths says in the footnotes of example 2.3 to just look it up in a reference book for now.

Although @julian 's post is quite nice.

 

[Rick16]

You should just be able to recognize them, look their values up in a table

That's why I decided to use an online calculator. I just thought that there might be a simple, straightforward way to solve this. Your demonstration with the zeta function is quite instructive and comes pretty close to the straightforward method that I was looking for, provided that I know the value of the zeta function.

 

[julian]

The particular derivation I gave requires knowledge of complex analysis and is quite lengthy. I therefore thought that it illustrated why deriving the result is not straightforward. However, I had forgotten that there is a substantially more elementary proof using Fourier series, which are covered in Griffiths.

Consider the Fourier series of $x^2$ on $-\pi < x < \pi$:

\begin{align*}
x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos(nx).
\end{align*}

Evaluating at $x = 0$:

\begin{align*}
0 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}.
\end{align*}

Therefore,

\begin{align*}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}.
\end{align*}

Splitting the alternating sum into odd and even parts:

\begin{align*}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}
& = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} - \sum_{n=1}^{\infty} \frac{1}{(2n)^2}
\nonumber \\
& = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} - \frac14 \sum_{n=1}^{\infty} \frac{1}{n^2}
\nonumber \\
& = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} - \frac14 \cdot \frac43 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}
\nonumber \\
& = \frac23 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}
\end{align*}

where we used $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac34 \sum_{n=1}^{\infty} \frac{1}{n^2}$ proven in earlier post. Hence:

\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8} .
\end{align*}

 

[Rick16]

However, I had forgotten that there is a substantially more elementary proof using Fourier series, which are covered in Griffiths.

Thanks a lot. This is a derivation that I can follow. For physics purposes it is of course much easier to look up the zeta function and tweek it so that it fits my series. But it is nice to see how this all comes about.

 

[MidgetDwarf]

For some of these, it is worth learning Persevals Identity (I know I spelled the name wrong).

 

[haruspex]

I don’t know if it can be formalised, but I once stumbled on this:
We know all the roots of sin(x), so write it as a factorised polynomial. We also know the series expansion. Equate the two and pair up the powers of x.

 

[WWGD]

TL;DR: I have a series and know its sum but I have no idea how to get there.

I have been working through problem 2.7 in Griffiths' QM book. Everything is pretty much clear, except that at the very end the following sum shows up: $(1/1 + 1/3^2 + 1/5^2 +1/7^2 +...)$ According to the solution, this sum equals $\pi^2/8$. How can this be determined?

1/n^2 has a known sum. Split into evens and odds.

 

[julian]

I don’t know if it can be formalised, but I once stumbled on this:
We know all the roots of sin(x), so write it as a factorised polynomial. We also know the series expansion. Equate the two and pair up the powers of x.

This is Euler's classic proof of the Basel problem that @.Scott was referring to in post #2. It appears in the book introduced by @mathwonk in another thread:



Euler’s factorisation argument for $sin⁡x$ was initially heuristic. A fully rigorous foundation came in the 19th century, especially through the theory of infinite products for analytic functions. In particular, see Weierstrass’s factorisation theorem.

For the benefit of @Rick16, here is Euler's heuristic proof:

The zeros of $\sin x$ are

\begin{align*}
0,\pm\pi,\pm2\pi,\pm3\pi,\ldots
\end{align*}

So, thinking of $\sin x/x$ as an "infinite-degree polynomial'', we are led to the factorisation

\begin{align*}
\frac{\sin x}{x}
&=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right) \cdots \\
&=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right).
\end{align*}

On the other hand, the Taylor series for $\sin x$ gives

\begin{align*}
\frac{\sin x}{x}
&=
1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\\
&=
1-\frac{x^2}{6}+\mathcal{O}(x^4).
\end{align*}

Expanding the product to order $x^2$,

\begin{align*}
\frac{\sin x}{x}
&=
\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right) \cdots
\\
&=
1-
\left(
\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots
\right)x^2
+\mathcal{O}(x^4)
\\
&=
1-
\left(
\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2}
\right)x^2
+\mathcal{O}(x^4)\\
&=
1-
\frac{x^2}{\pi^2}
\sum_{n=1}^{\infty}\frac{1}{n^2}
+\mathcal{O} (x^4).
\end{align*}

Comparing the coefficients of $x^2$,

\begin{align*}
\sum_{n=1}^{\infty}\frac{1}{n^2}
&=
\frac{\pi^2}{6}.
\end{align*}

 

[julian]

We can obtain the relevant sum directly by considering the factorisation of $\cos x$ instead:

\begin{align*}
\cos x
&=\left(1-\frac{x^2}{\pi^2/4}\right)\left(1-\frac{x^2}{9\pi^2/4}\right)\left(1-\frac{x^2}{25\pi^2/4}\right) \cdots \\
&=\prod_{n=0}^{\infty}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right).
\end{align*}

On the other hand, the Taylor series for $\cos x$ gives

\begin{align*}
\cos x
&=
1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\\
&=
1-\frac{x^2}{2}+\mathcal{O}(x^4).
\end{align*}

Expanding the product to order $x^2$,

\begin{align*}
\cos x
&=
\left(1-\frac{4x^2}{\pi^2}\right)\left(1-\frac{4x^2}{9\pi^2}\right)\left(1-\frac{4x^2}{25\pi^2}\right) \cdots
\\
&=
1-
\left(
\frac{1}{\pi^2} + \frac{1}{9\pi^2} + \frac{1}{25\pi^2} + \cdots
\right) 4 x^2
+\mathcal{O}(x^4)
\\
&=
1-
\frac{4 x^2}{\pi^2}
\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}
+\mathcal{O} (x^4).
\end{align*}

Comparing the coefficients of $x^2$,

\begin{align*}
\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}
&=
\frac{\pi^2}{8}.
\end{align*}

 

[Herman Trivilino]

I obviously don't need it in this case. I will need it for similar questions that may come up in the future.

Why use an online calculator when you can use your calculator. Notice that the terms in the series get really small rapidly. Add only the first few terms and the sum will be very close to $\pi^2/8 \approx 1.234$.

 

[julian]

@Rick16 asked how can the sum $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots$be determined?

In the following proof, the auxiliary function may not be an obvious choice, but the calculations are straightforward. The proof uses familiar differentiation and integration rules, together with partial differentiation and differentiation under the integral sign—techniques usually introduced at university.

We have

\begin{align*}
\int_0^1 \frac{\ln x}{x^2-1} dx & = - \int_0^\infty \frac{u e^{-u}}{e^{-2u}-1} du \qquad \qquad (x=e^{-u})
\nonumber \\
& = \int_0^\infty \frac{u}{e^u-1} du - \int_0^\infty \frac{u}{e^{2u}-1} du
\nonumber \\
& = \int_0^\infty \sum_{n=0}^\infty u e^{-(2n+1) u} du
\nonumber \\
& = \sum_{n=0}^\infty \int_0^\infty u e^{-(2n+1) u} du
\nonumber \\
& = \sum_{n=0}^\infty \frac{1}{(2n+1)^2} \int_0^\infty v e^{-v} dv \qquad \qquad (v=(2n+1)u)
\nonumber \\
& = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}
\end{align*}

The integral can be evaluated by elementary means using Feynman’s method of differentiation under the integral sign.

Define

\begin{align*}
I(\theta)
&= \int_0^1
\frac{\ln\left(1+(x^2-1)\sin^2\theta\right)}
{x^2-1}\,dx.
\end{align*}

We first differentiate with respect to $\theta$:

\begin{align*}
\frac{dI}{d\theta}
&=
\int_0^1
\frac{1}{x^2-1}
\frac{2(x^2-1)\sin\theta\cos\theta}
{1+(x^2-1)\sin^2\theta}\,dx\\
&=
\int_0^1
\frac{2 \sin\theta \cos\theta}
{\cos^2\theta+x^2\sin^2\theta}\,dx.
\end{align*}

For $0<\theta<\frac{\pi}{2}$,

\begin{align*}
\frac{dI}{d\theta}
&=
2 \sin\theta \cos\theta
\int_0^1
\frac{dx}
{\cos^2\theta+x^2\sin^2\theta}\\
&=
2 \sin\theta \cos\theta
\frac{1}{\sin\theta\cos\theta}
\tan^{-1}(\tan\theta)\\
&=2\theta.
\end{align*}

Now, since

\begin{align*}
I(0)
&=
\int_0^1
\frac{\ln 1}{x^2-1}\,dx
=0,
\end{align*}

we obtain

\begin{align*}
\int_0^1 \frac{\ln (x^2)}{x^2-1} dx
&=
I(\pi / 2) - I(0)\\
&=
\int_0^{\frac{\pi}{2}} \frac{dI}{d\theta}\,d\theta\\
&=
\int_0^{\frac{\pi}{2}} 2\theta\,d\theta\\
&=\left( {\frac{\pi}{2}} \right)^2.
\end{align*}