Question

Let $S_{n}=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots \ldots .$ to $n^{\text {th }}$ term. Then $\lim\limits _{n \rightarrow \infty} S_{n}$ is

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

Answer 1

Answer: (B)

$t_{n}=\cot ^{-1} 2 n^{2}$
$=\tan ^{-1} \frac{1}{2 n^{2}}=\tan ^{-1} \frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)}$
$=\tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)$
$\therefore S_{n}=\tan ^{-1}(2 n+1)-\tan ^{-1} 1$
$\therefore \lim\limits_{n \rightarrow \infty} S_{n}=\pi / 2-\pi / 4=\pi / 4$