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

Question

This is a homework for my son, he needs the proving.I tried to solve it by residue theory but I couldn't.

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

Answer 1

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*}

Answer 2

Another way to evaluate the sum is to use this. Note that $$ 576n^2 - 576n + 95=24^2(n^2-n+\frac{95}{24^2})=24^2[(n-\frac{1}{2})^2+(\frac{7i}{24})^2] $$ and $$ 144n^2 - 144n + 35=12^2(n^2-n+\frac{35}{12^2})=12^2[(n-\frac{1}{2})^2+(\frac{i}{12})^2] $$ and hence we have \begin{eqnarray*} &&\sum_{n=1}^\infty \left(\frac{14}{576n^2 - 576 + 95}-\frac{1}{144n^2 - 144n + 35}\right)\\ &=&\sum_{n=1}^\infty\left(\frac{7}{288}\frac{1}{(n-\frac{1}{2})^2+(\frac{7i}{24})^2}-\frac{1}{12^2}\frac{1}{(n-\frac{1}{2})^2+(\frac{i}{12})^2}\right) \\ &=& \frac{1}{2}\sum_{n=-\infty}^\infty\left(\frac{7}{288}\frac{1}{(n-\frac{1}{2})^2+(\frac{7i}{24})^2}-\frac{1}{12^2}\frac{1}{(n-\frac{1}{2})^2+(\frac{i}{12})^2}\right) \\ &=& \frac{1}{2} \left(\frac{\pi\sinh(2\pi b)}{b(\cosh(2\pi b)-\cos(2\pi a)}\bigg|_{a=\frac{1}{2},b=\frac{7i}{24}} - \frac{\pi\sinh(2\pi b)}{b(\cosh(2\pi b)-\cos(2\pi a)}\bigg|_{a=\frac{1}{2},b=\frac{i}{12}}\right)\\ &=&\frac{1}{2}\left(\frac{\pi(1+\sqrt3)}{12+24\sqrt2-12\sqrt3}-\frac{\pi}{24+12\sqrt3}\right)\\ &=&\frac{\pi}{24}(\sqrt6-\sqrt2). \end{eqnarray*}

Answer 3

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 $$-\left(\frac{\pi\cot\frac{19}{24}\pi}{14}-\frac{\pi\cot\frac7{12}\pi}{24}\right)=\frac{\pi}{24}\left(\cot\tfrac{7}{12}\pi-\cot\tfrac{19}{24}\pi\right)=\ldots=\frac{\pi}{24}(\sqrt6-\sqrt2).$$