TITLE: How to prove $\sum_{n=1}^{\infty}\frac{1}{16n^2-16n+3}=\frac{\pi}{8}$
QUESTION [6 upvotes]: Wolfram alpha computes 

$$\sum_{n=1}^{\infty}\frac{1}{16n^2-16n+3}=\frac{\pi}{8}$$

But I don't have any idea to prove this.
Thank you.

REPLY [6 votes]: $$\begin{align}
\sum_{n=1}^{\infty}\frac{1}{16n^2-16n+3}
&=\frac18\sum_{n=1}^{\infty}\left(\frac{1}{n-\frac34}-\frac1{n-\frac14}\right)\tag{1}\\
&=\frac18\sum_{n=1}^{\infty}\int_{0}^{1}\left(x^{n-7/4}-x^{n-5/4}\right)\,\mathrm dx\tag{2}\\
&=\frac18\int_{0}^{1}\sum_{n=1}^{\infty}\left(x^{n-7/4}-x^{n-5/4}\right)\,\mathrm dx\tag{3}\\
&=\frac18\int_{0}^{1}\left(\frac{x^{-3/4}}{1-x}-\frac{x^{-1/4}}{1-x}\right)\,\mathrm dx\tag{4}\\
&=\frac18\int_{0}^{1}\left(\frac{x^{-3/4}-x^{-1/4}}{1-x}\right)\,\mathrm dx\tag{5}\\
&=\frac18\int_{0}^{1}\frac{1}{\left(\sqrt{x}+1\right) x^{3/4}}\,\mathrm dx\tag{6}\\
&=\frac18\Bigg[4\arctan \left(\sqrt[4]{x}\right)\Bigg]_{0}^{1}\tag{7}\\
&=\frac\pi8\tag{8}\\
\end{align}$$

$$\large \sum_{n=1}^{\infty}\frac{1}{16n^2-16n+3}=\frac\pi8$$