Question

Let \(S_{n}=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots \ldots .\) to \(n^{\text {th }}\) term. Then \(\lim _{n \rightarrow \infty} S_{n}\) is
(A) \(\frac{\pi}{3}\)
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{6}\)
(D) \(\frac{\pi}{8}\)

Answer 1

Ans: (B)
Hint \(: t_{n}=\cot ^{-1} 2 n^{2}\)
$$
\begin{aligned}
&=\tan ^{-1} \frac{1}{2 n^{2}}=\tan ^{-1} \frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)} \\
&=\tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1) \\
&\therefore S_{n}=\tan ^{-1}(2 n+1)-\tan ^{-1} 1 \\
&\therefore \operatorname{Lim}_{n \rightarrow \infty} S_{n}=\pi / 2-\pi / 4=\pi / 4
\end{aligned}
$$