I’m working on the question below.
Find the value of $\sum_{k=0}^\infty \frac{1}{(2k + 1)^2}$ by adopting Parseval's identity for the function
$$f(x) = \begin{cases} -1 & \text{if } -\pi < x < 0 \\ 1 & \text{if }0 < x < \pi \\ 0 & \text{if }x = 0.\end{cases}$$
I've already got the Fourier series:
$$f(x) = \sum_{n=-\infty}^\infty \frac{1 -(-1)^n}{in\pi} e^{inx}.$$
So, I think equation of Parseval's identity is
$$\sum_{n=-\infty}^\infty \left(\frac{1-(-1)^n}{in\pi}\right)^2 = \frac{1}{2\pi}\int_{-\pi}^\pi |f(x)|^2 \; dx.$$
Is this ok?
But, I'm not sure how to conclude.
(Where does (2k+1) appear from this equation ?)
Best Answer
HINT:
Note that $1-(-1)^n=0$ when $n$ is even and $1-(-1)^n=2$ when $n$ is odd.
And you need to take the magnitude squared of the terms of the series, not their squares. So $|\frac1i|^2=1$.
Finally, account for the symmetry when summing $n$ from $-\infty$ to $\infty$.
The result will be $$\displaystyle 2\sum_{k=0}^\infty \frac4{\pi^2(2k+1)^2}=1\implies \sum_{k=0}^\infty \frac1{(2k+1)^2}=\frac{\pi^2}{8}$$