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So I have been solving various problems on Fourier series and this particular one got me struggling a bit.

Given a function $f(x) = |x|$ find a Fourier series on $[-\pi,\pi]$ and find the sum of following series: $\sum_{0}^\infty{\frac{1}{(2n-1)^2}}$.

So I successfully found a Fourier series of given function which is: $f(x) = \frac{\pi}{2} - \sum_{0}^{\infty}{\frac{1}{(2n+1)^2}}\pi\cos{nx}$. I have no problem of finding sum of $\frac{1}{(2n+1)^2}$ but how could I find the sum of $\sum_{0}^\infty{\frac{1}{(2n-1)^2}}$ ? I googled before posting this and all I could find is Parseval identity but that's not something I am familiar with. Can someone help me finding the sum or at least explain how could I do it with/without Parseval identity?

K.defaoite
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Filip Vlaisavljević
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  • Should by $\displaystyle f(x)\sim \frac{\pi}{2} - \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2} \cos (2n-1) x.$ – mjw Feb 16 '21 at 20:45

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$$f(x)=|x| \sim \frac{\pi}{2} - \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2}\cos (2n-1) x.$$

Evaluating this at $x=0$:

$$\frac{\pi}{2} = \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2}$$

$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$$

$$\sum_{n=0}^\infty \frac{1}{(2n-1)^2}=1+\frac{\pi^2}{8}$$

mjw
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