Question

The sum to infinite terms of the series ${\sin }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\left(\frac{\sqrt{2}-\sqrt{1}}{\sqrt{6}}\right)+{\sin }^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\right)+......\infty$ is equal to

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

Answer: (C)

$\frac{\pi }{2}$

Here, limit of the summation is from $1$ to $r$
$S_{\infty }={sin}^{-1}\frac{1}{\sqrt{2}}+{sin}^{-1}\left(\frac{\sqrt{2}-\sqrt{1}}{\sqrt{6}}\right)+{sin}^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}\right)={lim}_{r\to \infty }\sum {sin}^{-1}\left(\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}\right)={lim}_{r\to \infty }\sum {sin}^{-1}(\frac{\sqrt{r}}{\sqrt{r(r+1)}}-\frac{\sqrt{r-1}}{\sqrt{r(r+1)}})={lim}_{r\to \infty }\sum {sin}^{-1}(\frac{1}{\sqrt{r}}\sqrt{1-\frac{1}{r+1}}-\frac{1}{\sqrt{r+1}}\sqrt{1-\frac{1}{r}})={lim}_{r\to \infty }\sum ({sin}^{-1}\frac{1}{\sqrt{r}}-{sin}^{-1}\frac{1}{\sqrt{r+1}})={lim}_{r\to \infty }({sin}^{-1}1-{sin}^{-1}\frac{1}{\sqrt{r+1}})={lim}_{n\to \infty }(\frac{\pi }{2}-{sin}^{-1}\frac{1}{\sqrt{r+1}})={lim}_{r\to \infty }\left({cos}^{-1}\frac{1}{\sqrt{r+1}}\right)={cos}^{-1}0=\frac{\pi }{2}$