Question

If $\underset{n\to \infty }{lim}\sum _{a=2}^n{\sin }^{-1}\left[\frac{\sqrt{(a^2+2a)}-\sqrt{(a-1)(a+1)}}{a(a+1)}\right]=\frac{k\pi }{120},$ then the value of $k$ is

Answer 1

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

Putting $\frac{1}{a}=x,\frac{1}{a+1}=y$
Let,
$L=\left[\frac{\sqrt{(a^2+2a)}-\sqrt{(a-1)(a+1)}}{a(a+1)}\right]$
$=xy[\sqrt{(a+1{)}^2-1}-\sqrt{a^2-1}]$
$=xy[\sqrt{\frac{1}{y^2}-1}-\sqrt{\frac{1}{x^2}-1}]$
$=x\sqrt{1-y^2}-y\sqrt{1-x^2}$

$\therefore T_a={\sin }^{-1}(x\sqrt{1-y^2}-y\sqrt{1-x^2})$
$={\sin }^{-1}x-{\sin }^{-1}y={\sin }^{-1}\frac{1}{a}-{\sin }^{-1}\frac{1}{a+1}$
Substituting $a=2,3,4,...,n$
$S_n={\sin }^{-1}\frac{1}{2}-{\sin }^{-1}\frac{1}{3}+{\sin }^{-1}\frac{1}{3}-{\sin }^{-1}\frac{1}{4}+{\sin }^{-1}\frac{1}{4}-{\sin }^{-1}\frac{1}{5}::+{\sin }^{-1}\frac{1}{n}-{\sin }^{-1}\frac{1}{n+1}$

$\underset{n\to \infty }{lim}{S}_n=\underset{n\to \infty }{lim}({\sin }^{-1}\frac{1}{2}-{\sin }^{-1}\frac{1}{n+1})$
$=\frac{\pi }{6}+0=\frac{\pi }{6}$

$\Rightarrow k=20$