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

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.

$$\frac{1}{16n^2-16n+3} = \frac{1}{(4n-1)(4n-3)} = \frac{1}{2}\left(\frac{1}{4n-3}-\frac1{4n-1}\right)$$
So $$\sum_{n=1}^\infty \frac{1}{16n^2-16n+3} = \frac{1}{2}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2k-1}$$
That inner series is a well-known series for $\frac{\pi}4$.
\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}
$$\large \sum_{n=1}^{\infty}\frac{1}{16n^2-16n+3}=\frac\pi8$$