First slide

Properties of ITF

Question

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

Moderate

A

$\frac{π}{4}$
B

$- \frac{π}{4}$
C

$- \frac{π}{2}$
D

$\frac{π}{2}$

Solution

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

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