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.

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


You can apply the residue theorem after a bit of playing with the sums:
\begin{align*}&\sum_{n=1}^\infty\frac{14}{576n^2-576n+95}-\sum_{n=1}^\infty\frac4{576n^2-576+140}=\\&\sum_{n=1}^\infty\left(\frac1{24n-19}-\frac1{24n-5}\right)-\sum_{n=1}^\infty\left(\frac1{24n-14}-\frac1{24n-10}\right)=\\&\sum_{n=1}^\infty\left(\frac1{24n-19}+\frac1{24(1{-}n)-19}\right)-\sum_{n=1}^\infty\left(\frac1{24n-14}+\frac1{24(1{-}n)-14}\right)=\\&\sum_{n=-\infty}^\infty\frac1{24n-19}-\sum_{n=-\infty}^\infty\frac1{24n-14}=\sum_{n=-\infty}^\infty\frac5{(24n-19)(24n-14)}\end{align*}
Now consider
$$\lim_{n\to\infty}\int_{\varphi_{n+1/2}}\frac{5\pi\cot\pi z}{(24z-19)(24z-14)}dz=0,$$
where $\varphi_{n+1/2}$ is the circle of radius $n{+}\small 1/2$. The sum of all residues of the integrated function is also $0$ and the residues at points of $\mathbb Z$ gives us the terms of the sum.
But there are $2$ more, the residues at $\frac{19}{24}$ and $\frac{7}{12}$ are $\frac{5\pi\cot\frac{19}{24}\pi}{24\cdot(19-14)}$ and $\frac{5\pi\cot\frac7{12}\pi}{(14-19)\cdot 24}$ respectively, so your sum is equal to
$$-\left(\frac{\pi\cot\frac{19}{24}\pi}{14}-\frac{\pi\cot\frac7{12}\pi}{24}\right)=\frac{\pi}{24}\left(\cot\tfrac{7}{12}\pi-\cot\tfrac{19}{24}\pi\right)=\ldots=\frac{\pi}{24}(\sqrt6-\sqrt2).$$
