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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 ?)

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1 Answer 1

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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 $$\displaystyle 2\sum_{k=0}^\infty \frac4{\pi^2(2k+1)^2}=1\implies \sum_{k=0}^\infty \frac1{(2k+1)^2}=\frac{\pi^2}{8}$$

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  • $\begingroup$ Thanks your your help! Your solution is very clear! I'm very happy to be able to solve it! I've been spending a lot of time for this problem. $\endgroup$
    – EdwardIN
    May 13, 2020 at 3:42
  • $\begingroup$ You're welcome. My pleasure. Pleased that this was helpful. $\endgroup$
    – Mark Viola
    May 13, 2020 at 3:53

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