First slide

Properties of ITF

Question

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

Moderate

A

$\frac{π}{4}$
B

tan^–12
C

tan^–13
D

none of these

Solution

Let $T_{n} = \cot^{- 1} (n^{2} + \frac{3}{4}) \Rightarrow T_{n} = \cot^{- 1} (\frac{4 n^{2} + 3}{4})$
$\Rightarrow T_{n} = \tan^{- 1} (\frac{4}{4 n^{2} + 3}) = \tan^{- 1} (\frac{1}{n^{2} + \frac{3}{4}})$
$\begin{array}{l} \Rightarrow T_{n} = \tan^{- 1} (\frac{1}{1 + (n^{2} - \frac{1}{4})}) \\ \Rightarrow T_{n} = \tan^{- 1} (\frac{1}{1 + (n + \frac{1}{2}) (n - \frac{1}{2})}] \\ \Rightarrow T_{n} = \tan^{- 1} (\frac{(n + \frac{1}{2}) - (n - \frac{1}{2})}{1 + (n + \frac{1}{2}) (n - \frac{1}{2})}] \\ \Rightarrow T_{n} = \tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} (n - \frac{1}{2}) \end{array}$
Now $S_{n} = T_{1} + T_{2} + T_{3} + \dots + T_{n}$
$\Rightarrow S_{n} = \tan^{- 1} (\frac{3}{2}) - \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{5}{2}) - \tan^{- 1} (\frac{3}{2}) + \tan^{- 1} (\frac{7}{2}) - \tan^{- 1} (\frac{5}{2}) + \dots + \tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} (n + \frac{1}{2})$
$\begin{matrix} ∴ S_{\infty} = \tan^{- 1} \infty - \tan^{- 1} (\frac{1}{2}) \\ = \frac{π}{2} - \tan^{- 1} (\frac{1}{2}) = \cot^{- 1} (\frac{1}{2}) \\ ∴ S_{\infty} = \tan^{- 1} 2 \end{matrix}$

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