Summation of $\arcsin $ series. What is $a $ if 
$$\sum _{n=1} ^{\infty} \arcsin \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n (n+1)}}\right) =\frac {\pi }{a} \,?$$
Attempt: What I tried is to convert the series to $\arctan$ and then convert it telescoping series. So in terms of $\arctan $ it becomes 
$$\arctan \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n}+\sqrt {n-1}}\right) $$
but now if I divide by $n$ it simplifies as $n\frac {\pi}4-\sum _1^{\infty} \arctan \left(\frac {\sqrt {n-1}}{\sqrt {n}}\right) $ but as $n$ is tending towards infinity it will lead to infinity which seems wrong. Also note that $a$ is an integer . Thanks!

Taking the principal branch of $\arcsin$ (with values in $\bigl[-\frac{\pi}{2}, \frac{\pi}{2}\bigr]$), we have
$$\tan\bigl(\arcsin s\bigr) = \frac{\sin \bigl(\arcsin s\bigr)}{\cos \bigl(\arcsin s\bigr)} = \frac{s}{\sqrt{1 - s^2}}.$$
With $s = \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}$, we get
\begin{align}
1 - s^2 &= 1 - \frac{(\sqrt{n} - \sqrt{n-1})^2}{n(n+1)}\\
&= \frac{n^2 + n - (n - 2\sqrt{n(n-1)} + n-1)}{n(n+1)}\\
&= \frac{n(n-1) + 2\sqrt{n(n-1)} + 1}{n(n+1)}\\
&= \frac{(1 + \sqrt{n(n-1)})^2}{n(n+1)},
\end{align}
and so
\begin{align}
\tan \biggl(\arcsin \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\biggr)
&= \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\cdot \frac{\sqrt{n(n+1)}}{1 + \sqrt{n(n-1)}} \\
&= \frac{\sqrt{n} - \sqrt{n-1}}{1 + \sqrt{n} \sqrt{n-1}} \\
&= \tan\bigl(\arctan \sqrt{n} - \arctan \sqrt{n-1}\bigr),
\end{align}
whence we obtain
$$\sum_{n = 1}^{\infty} \arcsin \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}} = \sum_{n = 1}^\infty \bigl( \arctan \sqrt{n} - \arctan \sqrt{n-1}\bigr) = \frac{\pi}{2}.$$
