WBJEE · Maths · Limits
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 & Solution
Correct Answer
(B) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & 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 & \operatorname{Lim}_{n \rightarrow…
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