The sum of the infinite series {tan ^{ - 1}}l | vedclass

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(B) The general term of the series is $T_r = {	an ^{ - 1}}\left( {\frac{{2r}}{{1 - {r^2} + {r^4}}}} ight)$.We can rewrite the argument as: $\frac{{

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$\frac{\pi}{2}$

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(B) The general term of the series is $T_r = {	an ^{ - 1}}\left( {\frac{{2r}}{{1 - {r^2} + {r^4}}}} ight)$.We can rewrite the argument as: $\frac{{2r}}{{1 + {r^4} - {r^2}}} = \frac{{\left( {{r^2} + r} ight) - \left( {{r^2} - r} ight)}}{{1 + \left( {{r^2} + r} ight)\left( {{r^2} - r} ight)}}$.Using the identity ${	an ^{ - 1}}x - {	an ^{ - 1}}y = {	an ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} ight)$,we get:$T_r = {	an ^{ - 1}}\left( {{r^2} + r} ight) - {	an ^{ - 1}}\left( {{r^2} - r} ight)$.The sum of the first $n$ terms is $S_n = \sum_{r=1}^n [ {	an ^{ - 1}}(r^2+r) - {	an ^{ - 1}}(r^2-r) ]$.This is a telescoping series:$S_n = ( {	an ^{ - 1}}2 - {	an ^{ - 1}}0 ) + ( {	an ^{ - 1}}6 - {	an ^{ - 1}}2 ) + \dots + ( {	an ^{ - 1}}(n^2+n) - {	an ^{ - 1}}(n^2-n) )$.$S_n = {	an ^{ - 1}}(n^2+n) - {	an ^{ - 1}}0$.Taking the limit as $n 	o \infty$:$S = \mathop {\lim }\limits_{n 	o \infty } {	an ^{ - 1}}(n^2+n) - 0 = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

The sum of the infinite series ${\tan ^{ - 1}}\left( {\frac{2}{{1 - {1^2} + {1^4}}}} \right) + {\tan ^{ - 1}}\left( {\frac{4}{{1 - {2^2} + {2^4}}}} \right) + {\tan ^{ - 1}}\left( {\frac{6}{{1 - {3^2} + {3^4}}}} \right) + \dots$ is

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