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Let $$\sin x=\frac{2}{\pi}-\frac{4}{\pi}\cdot\sum_{n\ge1}\frac{\cos (2nx)}{4n^2-1}$$

be fourier series on $x\in\left[ 0,\pi\right]$. Show that

$$\frac{1}{1^2\cdot3^2}+\frac{1}{3^2\cdot5^2}+\frac{1}{5^2\cdot7^2}+...=\frac{\pi^2-8}{16}$$

I try to solve with partial fraction $$\frac{1}{(4n^2-1)^2}=\frac{1}{4}\cdot\left(\boxed{\frac{1}{2n+1}-\frac{1}{2n-1}}+\boxed{\frac{1}{(2n-1)^2}+\frac{1}{(2n+1)^2}}\right)$$

I think 1st box is telescopic series and using the sinx series to calculate the 2nd box.

I can solve the value of series with other technique e.g. $\zeta(2)$, Parseval's Identity, etc, But I would like to solve the series with fourier series of $\sin x$.

Any help would be appreciated.

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    Parseval's theorem would be useful. – 2017-02-17
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    @DanielFischer: You're fast. – 2017-02-17
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    Can I solve without Parseval's identity? The problem is given only fourier series of sinx and prove the statement by the given, Can I prove without the identity? – 2017-02-17
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    I have solved with Parseval' identity. I consider $f:=\frac{1}{2}-\frac{\pi}{4}\cdot\sin(\frac{x}{2})$ and use the identity to solve but I think It can solved only using fourier series of sinx. – 2017-02-17
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    Parseval's theorem is using Fourier series. Modulo constants, you want to evaluate the sum of the squares of the Fourier coefficients. That's precisely what Parseval's theorem is about. – 2017-02-17
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    Since you did not accept nor upvoted my answer, I realize that it wasn't helpful to you. I apologize for that. Hence I deleted it. – 2017-02-21

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