using Parseval's identity to estimate the value of $\Sigma_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}$ There is a problem with two parts; The first part is asking to find Fourier series for $f(x)=|\sin(x)|$ on $[-\pi,\pi]$. And the second part wants to estimate the following using Parseval's identity:
$$
\sum_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}
$$
I'm not sure it is necessary to use the answer of the first part or not. I don't know what to do!

The first part answer:

 $$\forall n, b_n=0$$.
$$
a_0=\frac{1}{\pi}\int_{-\pi}^\pi |\sin(x)|dx=\frac{2}{\pi}\int_0^\pi \sin(x)dx=\frac{-2}{\pi}\cos(x)]_0^\pi=\frac{-2}{\pi}(-1-1)=\frac{4}{\pi}
$$
\begin{align*}
a_n=\frac{1}{\pi}\int_{-\pi}^\pi |\sin(x)|\cos(nx) dx
= \frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)dx = \frac{1}{\pi}\int_0^\pi [\sin(x+nx)+\sin(x-nx)]\\ =\frac{1}{\pi}[\frac{-\cos((n+1)x)}{2(n+1)}+\frac{-\cos((1-n)x)}{2(1-n)}]_0^\pi \\
a_n=\frac{-1}{\pi}(\frac{\cos(n\pi)+1}{n^2-1})
\end{align*}

I guess I did it! 
I want to know the value of:
$$
\sum_{n=1}^\infty \frac{1}{(4n^2-1)^2}=\sum_{n=1}^\infty \frac{1}{((2n)^2-1)^2}
$$
So let $m=2n$, then $\sum_{n=1}^\infty \frac{1}{(m^2-1)^2}$. 
From the first part and Parseval's identity,
$$
\frac{2}{\pi}\int_0^\pi \sin^2(x)dx=\frac{16}{2\pi^2}\sum_{m=1}^\infty \frac{1}{\pi}\frac{((-1)^m+1)^2}{(m^2-1)^2}
$$
since $m=2n$,
$$
\sum_{m=1}^\infty \frac{1}{(m^2-1)^2}=\frac{\pi^3}{32}
$$
Is this correct?
