TITLE: Proving this formula $1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}=\sqrt2$
QUESTION [9 upvotes]: I tried to prove this formula but I couldn't do. 

$$1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}=\sqrt{2}$$

REPLY [2 votes]: $$\begin{align}
1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}
&=1+\frac1{4\pi}\sum_{n=0}^\infty\frac1{\left(n+\frac38\right)\left(n+\frac58\right)}\tag{1}\\
&=1+\frac1\pi\sum_{n=0}^\infty\left(\frac1{n+\frac38}-\frac1{n+\frac58}\right)\tag{2}\\
&=1+\frac1\pi\sum_{n=0}^\infty\int_0^1\left(x^{n-5/8}-x^{n-3/8}\right)\,\mathrm dx\tag{3}\\
&=1+\frac1\pi\int_0^1\left(\frac{x^{-5/8}-x^{-3/8}}{1-x}\right)\,\mathrm dx\tag{4}\\
&=1+\frac{1}{\pi} \Big[\pi\Big(\sqrt2-1\Big)\Big]\tag{5}\\
&=1+\sqrt2-1\tag{6}\\
&=\sqrt2\tag{7}\\
\end{align}$$

$$\large1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}
=\sqrt2
$$

I can add explanations if needed