Deduce that $\sum_{k \in \mathbb{N}} \frac{k^2}{(4k^2-1)^2} = \frac{\pi^2}{64}$

Question

Deduce that \begin{equation} \sum_{k \in \mathbb{N}} \frac{k^2}{(4k^2-1)^2} = \frac{\pi^2}{64} \end{equation} Given that f is a function $f: (0,\pi) → \mathbb{R}$ defined by $ x \to \cos(x)$.

I have already found the Fourier series of the odd extension of $f$ (which was the first part of the question) and i found it to be

\begin{equation} \sum_{n=2}^{\infty} \frac{2n(1+(-1)^n}{\pi(n^2-1)}. \end{equation}

I thought of using Parseval's Theorem however i dont seem to be getting anywhere with it. I would really appreciate some guidance.

Answer 1

Hint You are missing the sins from your series.

Split your sum in n odd and even.

\begin{equation} \sum_{n=2}^{\infty} \frac{2n(1+(-1)^n}{\pi(n^2-1)}=\left(\sum_{k=1}^{\infty} \frac{4k\left(1+(-1)^{2k}\right)}{\pi(4k^2-1)}\right)+\left(\sum_{k=1}^{\infty} \frac{2(2k+1)0}{\pi((2k+1)^2-1)}\right)\\=\sum_{k=1}^{\infty} \frac{8k}{\pi(4k^2-1)} \end{equation}

Now use parseval.

Answer 2

An answer has already helped you with Fourier analysis. If you're interested, here is a technique that uses complex analysis. Because your $a_n$ summand is even, use the fact that $\sum_{\mathbb{N}}a_n=-a_0/2+\frac{1}{2}\sum_{\mathbb{Z}}a_n$. In this case, $a_0=0$.

By the residue theorem, the sum over $\mathbb{Z}$ is just the sum of the residues of $-\pi\cot(\pi z)\frac{z^2}{(4z^2-1)^2}$ at $z=\pm 1/2$, which is $\pi^2/32$.

Your sum is thus $$\sum_{k\ge 1} \frac{k^2}{(4k-1)^2}=\frac{\pi^2}{64}$$