Find the sum : $\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$ Problem : 
Find the sum of : 
$$\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$$
My approach : 
Here the $n$'th term is given by : 
$$t_n = \sin^{-1}\left[\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right]$$
From now how to proceed further please suggest thanks....
 A: Hint: Use the fact that $\sin^{-1}a+\sin^{-1}b = \sin^{-1}(a\sqrt{1-b^2}+b\sqrt{1-a^2})$. The first two terms then give you $\sin^{-1}(\sqrt{2/3})$. Then applying the same identity with this term and the third term gives you $\sin^{-1}(\sqrt{3/4})$, and...
Edit: Induction step gives some headache, so let me write it for completeness. Suppose the sum of the first $n$ terms are $a = \sin^{-1}\sqrt\frac{n}{n+1}$. We show that the sum of the first $n+1$ terms are $\sin^{-1}\sqrt{\frac{n+1}{n+2}}$. Let $$b = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}$$ denote the $(n+1)$th term. Then, it is sufficient to prove that $c = a\sqrt{1-b^2}+b\sqrt{1-a^2} = \sqrt\frac{n+1}{n+2}$. Indeed, we have
\begin{align}
c &= \sqrt\frac{n}{n+1}\sqrt{1-\frac{(\sqrt{n+1}-\sqrt{n})^2}{(n+1)(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}\sqrt{1-\frac{n}{n+1}}\\
& = \sqrt\frac{n}{n+1}\sqrt{\frac{n^2+n+1+2\sqrt{n(n+1)}}{(n+1)(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)\sqrt{(n+2)}}
\end{align}
Noting that the nested radical $\sqrt{n^2+n+1+2\sqrt{n(n+1)}}$ is equal to $1+\sqrt{n(n+1)}$, we obtain
$$c = \sqrt n \frac{1+\sqrt{n(n+1)}}{(n+1)\sqrt{(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}} {(n+1)\sqrt{(n+2)}} = \sqrt\frac{n+1}{n+2}$$.
A: This is not an independent answer but a response to Lord Soth's speculation of a geometric proof.

Consider two right-handed triangles $ABC$ and $ABD$ with base $1$ and heights $\sqrt{n}$ and $\sqrt{n-1}$. Let $\theta$ be the angle $\measuredangle CBD$. The area of the triangle $BCD$ can be computed in two ways:


*

*$\frac12 |BC||BD| \sin\theta = \frac12 \sqrt{n+1} \sqrt{n} \sin\theta$

*$\frac12 |CD||AB| = \frac12 ( \sqrt{n} - \sqrt{n-1} )$
Equate them gives us:
$$\sin \theta = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}$$
On the other hand,
$$\begin{align}
\theta &= \measuredangle CBD = \measuredangle CBA - \measuredangle DBA\\
       &= \sin^{-1}\frac{|AC|}{|BC|} - \sin^{-1}\frac{|AD|}{|BD|}
        = \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}}
\end{align}$$
We obtain 
$$\sin^{-1}\left( \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right)
= \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}}$$
and we have turned the original series into a telescoping one.
The partial sum of the first $n$ terms of original series becomes the
angle $\measuredangle ABC$.
When $n \to \infty$, the line $BC$ becomes vertical and this geometrically justify why
the limit of the series is $\frac{\pi}{2}$.
