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 denominator as $1 + {

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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 denominator as $1 + {r^4} - {r^2} = 1 + {r^2}({r^2} - 1) = 1 + ({r^2} + r)({r^2} - r)$.Thus,$T_r = {	an ^{ - 1}}\left( {\frac{{({r^2} + r) - ({r^2} - r)}}{{1 + ({r^2} + r)({r^2} - r)}}} ight)$.Using the formula ${	an ^{ - 1}}x - {	an ^{ - 1}}y = {	an ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} ight)$,we get $T_r = {	an ^{ - 1}}({r^2} + r) - {	an ^{ - 1}}({r^2} - r)$.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:$S_n = ( {	an ^{ - 1}}2 - {	an ^{ - 1}}0 ) + ( {	an ^{ - 1}}6 - {	an ^{ - 1}}2 ) + \dots + ( {	an ^{ - 1}}({n^2} + n) - {	an ^{ - 1}}({n^2} - n) )$.$S_n = {	an ^{ - 1}}({n^2} + n) - {	an ^{ - 1}}0 = {	an ^{ - 1}}({n^2} + n)$.As $n 	o \infty$,$S_n 	o {	an ^{ - 1}}(\infty) = \frac{\pi}{2}$.

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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