Q. The sum of the infinite series $\sin^{- 1}{\left( \frac{1}{\sqrt{2}} \right) + \sin^{- 1}\left( \frac{\sqrt{2} - 1}{\sqrt{6}} \right)} + \sin^{- 1}{\left( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{12}} \right) + ...}$ $+ ... + \sin^{- 1}{\left( \frac{\sqrt{n} - \sqrt{(n - 1)}}{\sqrt{\{ n(n + 1)\}}} \right) + ...}$ is
A) $\frac{\pi}{8}$
B) $\frac{\pi}{4}$
C) $\frac{\pi}{2}$
D) $\pi$
Explanation
(c)
$\because T_{r} = \sin^{- 1}\left( \frac{\sqrt{r} - \sqrt{(r - 1)}}{\sqrt{r(r + 1)}} \right)$
$= \tan^{- 1}\left( \frac{\sqrt{r} - \sqrt{(r - 1)}}{1 + \sqrt{r}\sqrt{(r - 1)}} \right)$
$S_{n} = \sum_{r = 1}^{n}{\tan^{- 1}\left( \frac{\sqrt{r} - \sqrt{r - 1}}{1 + \sqrt{r}\sqrt{r - 1}} \right)}$
$= \sum_{r = 1}^{n}{\{\tan^{- 1}{\sqrt{r} - \tan^{- 1}\sqrt{(r - 1)}}}\ \}$
$= \tan^{- 1}{\sqrt{n} - \tan^{- 1}\sqrt{0}}$
$= \tan^{- 1}{\sqrt{n} - 0}$
$\therefore S_{\infty} = \tan^{- 1}{\infty = \frac{\pi}{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.