I want help to Prove $\pi \tanh(\frac{\pi }{2})=\sum_{n=0}^{\infty }\frac{1}{n^2+n+0.5}$ It is difficult, so I need how to start to prove it 
 $$\pi \tanh(\frac{\pi }{2})=\sum_{n=0}^{\infty }\frac{1}{n^2+n+0.5}$$

I am just mimicking the answer of @robjohn for this question, all credits go to him
let $x^2 + x + 0.5 = (x-a)(x-b)$, then we have $a+b = -1$, i.e. $b = -1-a$
\begin{align}
&\sum_{n=0}^{+\infty}\dfrac{1}{n^2 + n + 0.5} \\
= &\sum_{n=0}^{+\infty}\dfrac{1}{(n-a)(n-b)}\\
= & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} - \dfrac{1}{n-b}\right)\\
= & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} - \dfrac{1}{n+ 1 + a}\right)\\
= & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} + \dfrac{1}{-n- 1 - a}\right)\\
= & \dfrac{1}{a-b}\sum_{n=-\infty}^{+\infty}\dfrac{1}{n-a}\\
\end{align}
And we use the fact $\sum_{n=-\infty}^\infty\frac{1}{n-a}=-\pi\cot(\pi a)$,
so we have $$\sum_{n=0}^{+\infty}\dfrac{1}{n^2 + n + 0.5} = \dfrac{1}{b-a}\pi\cot(\pi a)$$
Now plug in the value of $a$ and $b$ to conclude.
