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

Question

Dr. Abubacker · Accepted Answer

(b)
Here, $T_{n} = \cot^{- 1}\left( n^{2} + \frac{3}{4} ight)$
$= 	an^{- 1}\left( \frac{4}{4n^{2} + 3} ight)$
=$	an^{- 1}\left( \frac{1}{1 + \left( n + \frac{1}{2} ight)\left( n - \frac{1}{2} ight)} ight)$
$= 	an^{- 1}\left\lbrack \frac{\left( n + \frac{1}{2} ight) - \left( n - \frac{1}{2} ight)}{1 + \left( n + \frac{1}{2} ight)\left( n - \frac{1}{2} ight)} ightbrack$
$= 	an^{- 1}{\left( n + \frac{1}{2} ight) - 	an^{- 1}\left( n - \frac{1}{2} ight)}$
$	herefore S_{\infty} = T_{\infty}^{- 1} - 	an^{- 1}\left( \frac{1}{2} ight)$
$= \frac{\pi}{2} - 	an^{- 1}\left( \frac{1}{2} ight)$
$\Rightarrow S_{\infty} = \cot^{- 1}\left( \frac{1}{2} ight)$
$\Rightarrow S_{\infty} = 	an^{- 1}{(2)}$

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

Explanation

Reviewed by Dr. Abubacker

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