# Prove the following series

1. Jun 5, 2014

### JasonHathaway

1. The problem statement, all variables and given/known data

Prove that, by putting x=0 x=∏ in $x^{2}=\frac{\pi^{2}}{3}+4 \sum\limits_{n=1}^\infty \frac{1}{n^{2}} cos(nx) cos(n \pi)$, that $\frac{\pi^{2}}{8}= \sum\limits_{n=1}^\infty \frac{1}{(2n+1)^{2}}$

3. The attempt at a solution

This a solved problem, I've understood the first two parts, and how the even elements of the series were eliminated, but what about $\frac{\pi^{2}}{6}$ and $\frac{\pi^{2}}{8}$?

Last edited: Jun 5, 2014
2. Jun 5, 2014

### Pranav-Arora

The answer is actually incorrect. The correct answer is $\frac{\pi^2}{8}-1$.

Rewrite series 2 in the following way:
$$\frac{\pi^2}{3}=4\left(\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots \right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots \right)\right)$$
$$\Rightarrow \frac{\pi^2}{3}=4\left(1+\sum_{n=1}^{\infty} \frac{1}{(2n+1)^2} -\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots \right)\right)\,\,\,\,\,\,(*)$$
From (3), you have $\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$, put this in $(*)$ and you should be able to reach the correct answer.

3. Jun 5, 2014

### Joffan

You can turn the equation (2) into an expression for $\frac{\pi^2}{12}$ easily enough, and then $\frac{\pi^2}{12}+\frac{\pi^2}{6}=\dots$

(and I disagree with Pranav-Arora - $\frac{\pi^2}{8}$ is correct)

Edit - Ah I see now what Pranav-Arora meant - the final summation should run from n=0, otherwise, indeed, the LHS should be $\frac{\pi^2}{8}-1$

Last edited: Jun 5, 2014