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!

We can utilize the following trigonometric identity 
$$\arcsin x-\arcsin y = \arcsin(x\sqrt{1-y^2}-y\sqrt{1-x^2})$$
by putting $x=\frac{1}{\sqrt{n}}$, $y=\frac{1}{\sqrt{n+1}}$. Then we get
\begin{align}\arcsin \frac{1}{\sqrt{n}}-\arcsin \frac{1}{\sqrt{n+1}} &= \arcsin\left(\frac{1}{\sqrt{n}}\sqrt{1-\frac{1}{n+1}}-\frac{1}{\sqrt{n+1}}\sqrt{1-\frac{1}{n}}\right)\\
&=\arcsin\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right). 
\end{align}
So we have a telescoping series
\begin{align}
\sum_{n=1}^{k} \arcsin\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right) &= \sum_{n=1}^{k} \left(\arcsin \frac{1}{\sqrt{n}}-\arcsin \frac{1}{\sqrt{n+1}}\right)\\
&=\arcsin 1-\arcsin \frac{1}{\sqrt{k+1}}
\end{align}
and in the limit $\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right)=\arcsin 1-\arcsin 0=\frac{\pi}{2}$ and $a=2$.
