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

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

 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\large 1
+\sum_{n\ =\ 0}^{\infty}{1 \over \pi\pars{2n + 3/4}\pars{2n + 5/4}}}
=1 + {1 \over 4\pi}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{n + 3/8}\pars{n + 5/8}}
\\[5mm]&=1 + {1 \over 4\pi}\bracks{\Psi\pars{3/8} - \Psi\pars{5/8} \over 3/8 - 5/8}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
Digamma Function $\color{#000}{\bf 6.3.1}$.

Then
  \begin{align}&\color{#66f}{\large 1
+\sum_{n\ =\ 0}^{\infty}{1 \over \pi\pars{2n + 3/4}\pars{2n + 5/4}}}
=1 + {1 \over \pi}\bracks{\Psi\pars{5 \over 8} - \Psi\pars{3 \over 8}}
\end{align}

With
Euler Reflection Formula $\color{#000}{\bf 6.3.7}$
$\ds{\pars{~\Psi\pars{1 - z} = \Psi\pars{z} + \pi\cot\pars{\pi z}~}}$ we'll get:
\begin{align}&\color{#66f}{\large 1
+\sum_{n\ =\ 0}^{\infty}{1 \over \pi\pars{2n + 3/4}\pars{2n + 5/4}}}
=1 + {1 \over \pi}\bracks{\pi\cot\pars{\pi\,{3 \over 8}}}
=1 + \cot\pars{3\pi \over 8}
\\[5mm]&=1 + {1 \over \tan\pars{135^{\circ}/2}}
=1 + {1 \over 1/\pars{\root{2} - 1}} =\color{#66f}{\large\root{2}}\,.\quad
{\tt\pars{~\mbox{See a proof below}~}}.
\end{align}

Also,
  \begin{align}
-1&=\tan\pars{135^{\circ}}
= {2\tan\pars{135^{\circ} /2} \over 1 - \tan^{2}\pars{135^{\circ} /2}}
\quad\imp\quad
\tan^{2}\pars{135^{\circ} \over 2} - 2\tan\pars{135^{\circ} \over 2} - 1 = 0
\\[5mm]&\imp\quad\tan\pars{135^{\circ} \over 2}
={2 + \sqrt{8} \over 2} = 1 + \root{2} = {1 \over \root{2} - 1}
\end{align}

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

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

I can add explanations if needed
A: $$
\begin{align}
1+\frac1{4\pi}\sum_{n=0}^\infty\frac1{(n+\frac38)(n+\frac58)}
&=1+\frac1\pi\sum_{n=0}^\infty\left(\frac1{n+\frac38}-\frac1{n+\frac58}\right)\tag{1}\\
&=1+\frac1\pi\sum_{n=0}^\infty\left(\vphantom{\frac1{n+\frac38}}\right.\underbrace{\frac1{n+\frac38}}_{n+\frac38\text{ for }n\ge0}+\underbrace{\frac1{-n-1+\frac38}}_{n+\frac38\text{ for }n\le-1}\left.\vphantom{\frac1{n+\frac38}}\right)\tag{2}\\
&=1+\frac1\pi\color{#C00000}{\sum_{n=-\infty}^\infty\frac1{n+\frac38}}\tag{3}\\
&=1+\frac1\pi\color{#C00000}{\pi\cot\left(\frac38\pi\right)}\tag{4}\\[6pt]
&=1+\sqrt2-1\tag{5}\\[12pt]
&=\sqrt2\tag{6}
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: rewrite terms and set up next step
$(3)$: write as a sum over $\mathbb{Z}$
$(4)$: use $\color{#C00000}{(7)}$ from this answer
$(5)$: evaluate $\cot\left(\frac38\pi\right)$  
