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](https://sagecell.sagemath.org/?z=eJwrTbI1NQACa67kjEyNPE3b6sw0jTxVQwNbW0OFmhoFCNNSJ9nWUAcuY4yQMQfK6BoCCQNNTeui1JLSojyNZE3rWq5i2-LSXI08oLbSJB2I2foaeXFGQGVcBUWZeSUaxVqGenBOQGackb6GqVZxYVGJhqkmUBUAw8Eq3Q==&lang=gp&interacts=eJyLjgUAARUAuQ==).



A conjectural infinite series for $\frac{\pi^2}{5\sqrt{5}}$
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 $S\_{10,1}-S\_{10,3}$.


