using Parseval's identity to estimate the value of $\Sigma_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}$

Question

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*}

Answer 1

The Fourier series of the function $f(x)$ is given by $$ f(x) = \sum_{n=-\infty}^\infty c_n e^{i n x} $$ with $$ c_n = \frac{1}{2 \pi} \int_{-\pi}^\pi d x~f(x) e^{-\imath n x} $$ In the case $f(x) = |\sin x |$, this results in $$ c_n = 2 \frac{1}{2 \pi} \int_{0}^\pi d x~\sin x e^{-\imath n x} = \left\{ \begin{array}{ll} -\frac{2}{\pi (n^2-1)} & \text{for $n$ even}\\ 0 & \text{for $n$ odd} \end{array} \right. $$ Using Parsifal's identity we have $$ \sum_{n=-\infty}^\infty |c_n|^2 = \frac{1}{2 \pi} \int_{-\pi}^\pi d x ~ |\sin x|^2 = \frac{1}{2} $$ Since $c_{n} = c_{-n}$ the sum on the left hand side can be rewritten as $$ c_0^2 + 2 \sum_{n=1}^{\infty} c_n^2 = \frac{1}{2} $$ From which it follows that with $c_0=\frac{2}{\pi}$ $$ \sum_{n=1}^{\infty} c_n^2 = \frac{1}{2}\left(\frac{1}{2} - c_0^2\right) = \frac{1}{4} - \frac{2}{\pi^2} $$ Using the result we found for $c_n$ we get $$ \sum_{n=1}^{\infty} c_n^2 = \sum_{n=1}^{\infty} c_{2 n - 1}^2 + \sum_{n=1}^{\infty} c_{2 n}^2 = \sum_{n=1}^{\infty} c_{2 n}^2 = \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{(4 n^2 -1)^2} $$ and therefore we finally obtain $$ \sum_{n=1}^{\infty} \frac{1}{(4 n^2 -1)^2} = \frac{\pi^2}{4} \left( \frac{1}{4} - \frac{2}{\pi^2} \right) = \frac{\pi^2-8}{16} $$

Answer 2

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?