Question

The sum of the infinite series
${\sin }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{6}}\right)+{\sin }^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}\right)+\cdots +{\sin }^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)+\cdots$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is C $\frac{\pi }{2}$

We have,
${\sin }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{6}}\right)+{\sin }^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}\right)+\cdots +{\sin }^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)$
Now, $T_n={\sin }^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)$
$={\sin }^{-1}(\frac{1}{\sqrt{n+1}}-\frac{\sqrt{n-1}}{\sqrt{n+1}\sqrt{n}})$
$={\sin }^{-1}[\frac{1}{\sqrt{n}}\sqrt{1-{\left(\frac{1}{\sqrt{n+1}}\right)}^2}-\frac{1}{\sqrt{n+1}}\sqrt{1-{\left(\frac{1}{\sqrt{n}}\right)}^2}]$
Using the formula:
$[{\sin }^{-1}x-{\sin }^{-1}y={\sin }^{-1}(x\sqrt{1-y^2}-y\sqrt{1-x^2})]$
$={\sin }^{-1}\frac{1}{\sqrt{n}}-{\sin }^{-1}\frac{1}{\sqrt{n+1}}$
$T_1={\sin }^{-1}1-{\sin }^{-1}\frac{1}{\sqrt{2}}$
$T_2={\sin }^{-1}\frac{1}{\sqrt{2}}-{\sin }^{-1}\frac{1}{\sqrt{3}}$
$\begin{array}{ccc}⋮& ⋮& ⋮\end{array}$
$T_n={\sin }^{-1}\frac{1}{\sqrt{n}}-{\sin }^{-1}\frac{1}{\sqrt{n+1}}$
Adding all the terms, we get
$S_n={\sin }^{-1}1-{\sin }^{-1}\frac{1}{\sqrt{n+1}}$
$\therefore S_{\infty }=\frac{\pi }{2}-0=\frac{\pi }{2}$