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.