Question

Let $S_n={\cot }^{-1}2+{\cot }^{-1}8+{\cot }^{-1}18+{\cot }^{-1}32+\dots \dots .$ to $n^{\text{th }}$ term. Then $\underset{n\to \infty }{\lim }{S}_n$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{8}$

Answer 1

Answer: (B)

$t_n={\cot }^{-1}2n^2$
$={\tan }^{-1}\frac{1}{2n^2}={\tan }^{-1}\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$
$={\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)$
$\therefore S_n={\tan }^{-1}(2n+1)-{\tan }^{-1}1$
$\therefore \underset{n\to \infty }{\lim }{S}_n=\pi /2-\pi /4=\pi /4$