The sum ∑n=1∞ tan−1⁡1n2+n+1 is equal to

∑n=1∞ tan−1⁡1n2+n+1=∑n=1∞ tan−1⁡(n+1)−(n)1+(n+1)(n)=∑n=1∞ tan−1 (n+1)-tan-1n=tan-12-tan-11+tan-13-tan-12+.......tan-1∞=π2−tan−1⁡1=π4

The sum ∑n=1∞ tan−1⁡1n2+n+1 is equal to

∑n=1∞ tan−1⁡1n2+n+1=∑n=1∞ tan−1⁡(n+1)−(n)1+(n+1)(n)=∑n=1∞ tan−1 (n+1)-tan-1n=tan-12-tan-11+tan-13-tan-12+.......tan-1∞=π2−tan−1⁡1=π4

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$The sum \sum_{n = 1}^{\infty} \tan^{- 1} (\frac{1}{n^{2} + n + 1}) is equal to$

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$\frac{π}{4}$

$- \frac{π}{4}$

$- \frac{π}{2}$

$\frac{π}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} \sum_{n = 1}^{\infty} \tan^{- 1} (\frac{1}{n^{2} + n + 1}) \\ = \sum_{n = 1}^{\infty} \tan^{- 1} \frac{(n + 1) - (n)}{1 + (n + 1) (n)} = \sum_{n = 1}^{\infty} \tan^{- 1} (n + 1) - \tan^{- 1} n \\ = \tan^{- 1} 2 - \tan^{- 1} 1 + \tan^{- 1} 3 - \tan^{- 1} 2 + . . . . . . . \tan^{- 1} \infty \\ = \frac{π}{2} - \tan^{- 1} 1 = \frac{π}{4} \end{array}$