Find the sum of the series: tan^{-1} left( fr | vedclass

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(A) The general term of the series is $T_k = 	an^{-1} \left( \frac{1}{1 + k(k + 1)} ight)$.We can rewrite the argument as $\frac{(k + 1) - k}{1 + k

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

Vedclass · Answer

(A) The general term of the series is $T_k = 	an^{-1} \left( \frac{1}{1 + k(k + 1)} ight)$.We can rewrite the argument as $\frac{(k + 1) - k}{1 + k(k + 1)}$.Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$, we get $T_k = 	an^{-1}(k + 1) - 	an^{-1}(k)$.The sum $S_n = \sum_{k=1}^{n} (	an^{-1}(k + 1) - 	an^{-1}(k))$.This is a telescoping series: $S_n = (	an^{-1} 2 - 	an^{-1} 1) + (	an^{-1} 3 - 	an^{-1} 2) + \dots + (	an^{-1}(n + 1) - 	an^{-1} n)$.All intermediate terms cancel out, leaving $S_n = 	an^{-1}(n + 1) - 	an^{-1}(1)$.Using the identity again, $S_n = 	an^{-1} \left( \frac{(n + 1) - 1}{1 + (n + 1)(1)} ight) = 	an^{-1} \left( \frac{n}{n + 2} ight)$.

Find the sum of the series: $\tan^{-1} \left( \frac{1}{1 + 1 \times 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \times 3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n(n + 1)} \right) =$

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