Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

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

Let $S={\sin }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin \frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+...+{\sin }^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)$

Now $T_n={\sin }^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)$

$={\sin }^{-1}[\frac{1}{\sqrt{n}}\sqrt{1-{\left(\frac{1}{\sqrt{n+1}}\right)}^2}-\frac{1}{\sqrt{n+1}}\sqrt{1-{\left(\frac{1}{\sqrt{n}}\right)}^2}]$

$={\sin }^{-1}\frac{1}{\sqrt{n}}-{\sin }^{-1}\frac{1}{\sqrt{n+1}}$

$\therefore S={\sin }^{-1}\frac{1}{\sqrt{2}}+\left({\sin }^{-1}\frac{1}{\sqrt{2}}-\sin \frac{1}{\sqrt{3}}\right)+({\sin }^{-1}\frac{1}{\sqrt{3}}-\sin \frac{1}{\sqrt{4}})+...+\infty$

$=2{\sin }^{-1}\frac{1}{\sqrt{2}}=2\left(\frac{\pi }{4}\right)=\frac{\pi }{2}$