Why is $$ \sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right) = \frac{\pi^2}{50} \frac{1}{1- \cos \frac{\pi}{5}} $$ ?
Each form of $$ \sum_{n=1}^{\infty} \frac{1}{(10n-9)^2} $$ and $$ \sum_{n=1}^{\infty} \frac{1}{(10n-1)^2} $$
couldn't be calculated but the sum of them has a closed form
The $\Gamma$ function fulfills the symmetry relation $$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} $$ hence by applying $\frac{d}{dz}\left(\log(\cdot)\right)$ to both sides $$\psi(z)-\psi(1-z) =-\pi\cot(\pi z) $$ and by differentiating once again $$ \psi'(z)+\psi'(1-z) = \frac{\pi^2}{\sin^2(\pi z)}. \tag{1}$$ Since $$ \psi'(z) = \sum_{n\geq 0}\frac{1}{(n+z)^2}\tag{2} $$ equations $(1)$ and $(2)$ with $z=\frac{1}{10}$ prove the claim.
Using $$\sum_{n\in\mathbb{Z}}\frac{1}{(n-a)^2+b^2}=\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}$$ from this, one has \begin{eqnarray} &&\sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right)\\ &=&\sum_{n\in\mathbb{Z}}\frac{1}{(10n-1)^2}=\frac{1}{10^2}\sum_{n\in\mathbb{Z}}\frac{1}{(n-\frac1{10})^2}\\ &=&\frac1{10}\lim_{b\to0}\frac{\pi\sinh(2\pi b)}{b\left(\cosh(2\pi b)-\cos(\frac{2\pi}{10})\right)}\\ &=&\frac{\pi^2}{50} \frac{1}{1- \cos \frac{\pi}{5}}. \end{eqnarray}
