I am looking for a proof of the following claim:
First define the function $\chi(n)$ as follows:
$$\chi(n)=\begin{cases}1, & \text{if }n \equiv \pm 1 \pmod{10} \\
-1, & \text{if }n \equiv \pm 3 \pmod{10} \\ 0, & \text{if otherwise }
\end{cases}$$
Then,
$$\frac{\pi^2}{5\sqrt{5}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n^2}$$
The SageMath cell that demonstrates this claim can be found here.
Best Answer
More generally, if $1\le k\le N-1$ is an integer, where $N$ is a positive interger, $$S_{N,k} := \sum_{n=0}^\infty\biggl( \frac{1}{(N n+N-k)^2} + \frac{1}{(N n+k)^2} \biggr) = \frac{\pi^2}{N^2\sin^2(\pi k/N)}.$$ Your sum is $S_{10,1}-S_{10,3}$.