Proving of $\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$ This is a homework for my son, he needs the proving.I tried to solve it by residue theory but I couldn't.

$$\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$$


I wouldn't say it's very illuminating, but here's an answer:
We have
\begin{align*}
\sum_{n=1}^\infty \frac{14}{576n^2 - 576 + 95}
  &= \sum_{n=1}^\infty \frac{1}{24n - 19} - \frac{1}{24n - 5} \\
  &= \frac{1}{24} \sum_{n=1}^\infty \frac{1}{n - 19/24} - \frac{1}{n - 5/24} \\
  &= \frac{1}{24} \left(\psi\left(\frac{19}{24}\right) - \psi\left(\frac{5}{24}\right)\right),
\end{align*}
where $\psi$ is the digamma function. (The last equality follows more or less directly from its definition, or at least one of its definitions.) Since $\psi$ happens to satisfy $\psi(1 - z) - \psi(z) = \pi \cot \pi z$ (cf. the corresponding formula for $\Gamma$), we therefore have
\begin{align*}
\sum_{n=1}^\infty \frac{14}{576n^2 - 576 + 95}
  &= \frac{\pi}{24} \cot \frac{5\pi}{24}
\end{align*}
The same method gives
\begin{align*}
\sum_{n=1}^\infty \frac{1}{144n^2 - 144n + 35}
  &= \frac{1}{2}\sum_{n=1}^\infty \frac{1}{12n - 7} - \frac{1}{12n - 5} \\
  &= \frac{1}{24} \left(\psi\left(\frac{7}{12}\right) - \psi\left(\frac{5}{12}\right)\right) \\
  &=  \frac{\pi}{24}\cot \frac{5\pi}{12}.
\end{align*}
Thus the given sum is equal to
\begin{align*}
\frac{\pi}{24}\left(\cot \frac{5\pi}{24} - \cot \frac{5\pi}{12}\right) = \frac{\pi}{24}\left(\sqrt{6} - \sqrt{2}\right).
\end{align*}
