The sum of the infinite series cot ^{-1}left( | vedclass

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(D) The general term of the series is $T_n = \cot^{-1}\left(\frac{4n^2+3}{4}ight) = 	an^{-1}\left(\frac{4}{4n^2+3}ight)$.We can rewrite this as $

Vedclass · Accepted Answer

$\frac{\pi}{2}-	an ^{-1}\left(\frac{1}{2}ight)$

Vedclass · Answer

(D) The general term of the series is $T_n = \cot^{-1}\left(\frac{4n^2+3}{4}ight) = 	an^{-1}\left(\frac{4}{4n^2+3}ight)$.We can rewrite this as $T_n = 	an^{-1}\left(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}ight)$.Using the formula $	an^{-1}x - 	an^{-1}y = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$,we get $T_n = 	an^{-1}(2n+1) - 	an^{-1}(2n-1)$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^n (	an^{-1}(2k+1) - 	an^{-1}(2k-1))$.This is a telescoping series: $S_n = 	an^{-1}(2n+1) - 	an^{-1}(1)$.As $n 	o \infty$,$S_{\infty} = \lim_{n 	o \infty} 	an^{-1}(2n+1) - 	an^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.Note that $	an^{-1}(1/2) = \frac{\pi}{2} - \cot^{-1}(1/2)$.Comparing with the options,$\frac{\pi}{2} - 	an^{-1}(2) = 	an^{-1}(1/2)$.Given the options,the correct form is $\frac{\pi}{2} - 	an^{-1}(2) = 	an^{-1}(1/2)$,which is equivalent to $\frac{\pi}{2} - \cot^{-1}(1/2)$.

The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \ldots$ is :-

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