Sum of infinite terms of the series cot−1⁡12+34+cot−1⁡22+34+cot−1⁡32+34+… is

sn=∑n=1n tn=tan−1⁡n+12−tan−1⁡12cot−1⁡12+34+cot−1⁡22+34+cot−1⁡32+34+…=∑n=1∞cot−1⁡n2+34=∑n=1∞tan−1⁡1n2+34=∑n=1∞tan−1⁡11+n2-14Let Sn=∑n=1ntan−1⁡11+n2-14Sn=∑n=1ntan−1⁡n+12-n-121+n+12n-12Sn=∑n=1ntan−1⁡n+12-tan−1⁡n-12Sn=tan−1⁡32-tan−1⁡12+tan−1⁡52-tan−1⁡32+.....+tan−1⁡n+12-tan−1⁡n-12Sn=tan−1⁡n+12-tan−1⁡12s∞=tan−1⁡(∞)−tan−1⁡12=π2−tan−1⁡12=cot−1⁡12=tan−1⁡2

Sum of infinite terms of the series cot−1⁡12+34+cot−1⁡22+34+cot−1⁡32+34+… is

sn=∑n=1n tn=tan−1⁡n+12−tan−1⁡12cot−1⁡12+34+cot−1⁡22+34+cot−1⁡32+34+…=∑n=1∞cot−1⁡n2+34=∑n=1∞tan−1⁡1n2+34=∑n=1∞tan−1⁡11+n2-14Let Sn=∑n=1ntan−1⁡11+n2-14Sn=∑n=1ntan−1⁡n+12-n-121+n+12n-12Sn=∑n=1ntan−1⁡n+12-tan−1⁡n-12Sn=tan−1⁡32-tan−1⁡12+tan−1⁡52-tan−1⁡32+.....+tan−1⁡n+12-tan−1⁡n-12Sn=tan−1⁡n+12-tan−1⁡12s∞=tan−1⁡(∞)−tan−1⁡12=π2−tan−1⁡12=cot−1⁡12=tan−1⁡2

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Sum of infinite terms of the series $\cot^{- 1} (1^{2} + \frac{3}{4}) + \cot^{- 1} (2^{2} + \frac{3}{4}) + \cot^{- 1} (3^{2} + \frac{3}{4}) + \dots$ is

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tan^–12

$\frac{π}{4}$

tan^–13

tan^–14

answer is B.

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Detailed Solution

$\begin{array}{l} s_{n} = \sum_{n = 1}^{n} t_{n} = \tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} \frac{1}{2} \\ \cot^{- 1} (1^{2} + \frac{3}{4}) + \cot^{- 1} (2^{2} + \frac{3}{4}) + \cot^{- 1} (3^{2} + \frac{3}{4}) + \dots \\ = \sum_{n = 1}^{\infty} \cot^{- 1} (n^{2} + \frac{3}{4}) = \sum_{n = 1}^{\infty} \tan^{- 1} \frac{1}{(n^{2} + \frac{3}{4})} = \sum_{n = 1}^{\infty} \tan^{- 1} \frac{1}{1 + (n^{2} - \frac{1}{4})} \\ L e t S_{n =} \sum_{n = 1}^{n} \tan^{- 1} \frac{1}{1 + (n^{2} - \frac{1}{4})} \\ S_{n} = \sum_{n = 1}^{n} \tan^{- 1} \frac{(n + \frac{1}{2}) - (n - \frac{1}{2})}{1 + (n + \frac{1}{2}) (n - \frac{1}{2})} \\ S_{n} = \sum_{n = 1}^{n} \tan^{- 1} (n + \frac{1}{2}) - t {an}^{- 1} (n - \frac{1}{2}) \\ S_{n} = \tan^{- 1} (\frac{3}{2}) - t {an}^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{5}{2}) - t {an}^{- 1} (\frac{3}{2}) + . . . . . + \tan^{- 1} (n + \frac{1}{2}) - t {an}^{- 1} (n - \frac{1}{2}) \\ S_{n} = \tan^{- 1} (n + \frac{1}{2}) - t {an}^{- 1} (\frac{1}{2}) \\ s_{\infty} = \tan^{- 1} (\infty) - \tan^{- 1} \frac{1}{2} = \frac{π}{2} - \tan^{- 1} \frac{1}{2} \\ = \cot^{- 1} \frac{1}{2} = \tan^{- 1} 2 \end{array}$