How Does Setting x=0 and x=π Simplify the Series to Prove a Summation Identity?

[JasonHathaway]

Homework Statement

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

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 [itex]\frac{\pi^{2}}{6} [/itex] and [itex]\frac{\pi^{2}}{8} [/itex]?

 

[Saitama]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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.

 

[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##