First slide

Properties of ITF

Question

The value of
$\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{13} + \dots + \tan^{- 1} \frac{1}{n^{2} + n + 1} + \dots$ to $\infty$ , is

Moderate

A

$\frac{π}{2}$
B

$\frac{π}{4}$
C

$\frac{2 π}{3}$
D

$0$

Solution

We have,
$\begin{array}{l} \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{13} + \dots + \tan^{- 1} \frac{1}{n^{2} + n + 1} + \dots . . to \infty \\ = lim_{n \to \infty} \sum_{r = 1}^{n} \tan^{- 1} \frac{1}{r^{2} + r + 1} \\ = lim_{n \to \infty} \sum_{r = 1}^{n} \tan^{- 1} \{\frac{1}{1 + r (r + 1)}\} \\ = lim_{n \to \infty} \sum_{r = 1}^{n} \tan^{- 1} \{\frac{(r + 1) - r}{1 + r (r + 1)}\} \\ = lim_{n \to \infty} \sum_{r = 1}^{n} \{\tan^{- 1} (r + 1) - \tan^{- 1} r\} \\ = lim_{n \to \infty} \{\tan^{- 1} (n + 1) - \tan^{- 1} 1\} \\ = \tan^{- 1} \infty - \tan^{- 1} 1 = \frac{π}{2} - \frac{π}{4} = \frac{π}{4} \end{array}$

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