The sum to infinite terms of the series tan^{ | vedclass

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Question

(A) The $n^{th}$ term of the series is $T_n = 	an^{-1}\left(\frac{2^{n-1}}{1+2^{2n-1}}ight)$.We can rewrite the argument as $\frac{2^n - 2^{n-1}}{1

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$\frac{\pi}{4}$

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(A) The $n^{th}$ term of the series is $T_n = 	an^{-1}\left(\frac{2^{n-1}}{1+2^{2n-1}}ight)$.We can rewrite the argument as $\frac{2^n - 2^{n-1}}{1 + 2^n \cdot 2^{n-1}}$.Using the formula $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$,we get $T_n = 	an^{-1}(2^n) - 	an^{-1}(2^{n-1})$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (	an^{-1}(2^k) - 	an^{-1}(2^{k-1}))$.This is a telescoping series: $S_n = (	an^{-1}(2^1) - 	an^{-1}(2^0)) + (	an^{-1}(2^2) - 	an^{-1}(2^1)) + \dots + (	an^{-1}(2^n) - 	an^{-1}(2^{n-1}))$.$S_n = 	an^{-1}(2^n) - 	an^{-1}(1) = 	an^{-1}(2^n) - \frac{\pi}{4}$.As $n 	o \infty$,$S_n = 	an^{-1}(\infty) - \frac{\pi}{4} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

The sum to infinite terms of the series $\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{2}{9}\right) + \dots + \tan^{-1}\left(\frac{2^{n-1}}{1+2^{2n-1}}\right) + \dots$ is

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