If f(n) = tan^{-1} left( frac{e-1}{e^{-n} + e | vedclass

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(D) Given $f(n) = 	an^{-1} \left( \frac{e-1}{e^{-n} + e^{n+1}} ight)$.Multiply the numerator and denominator by $e^n$:$f(n) = 	an^{-1} \left( \fra

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$tan^{-1}(\frac{1}{e})$

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(D) Given $f(n) = 	an^{-1} \left( \frac{e-1}{e^{-n} + e^{n+1}} ight)$.Multiply the numerator and denominator by $e^n$:$f(n) = 	an^{-1} \left( \frac{e^n(e-1)}{1 + e^{2n+1}} ight) = 	an^{-1} \left( \frac{e^{n+1} - e^n}{1 + e^{n+1} \cdot e^n} ight)$.Using the identity $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$,we get:$f(n) = 	an^{-1}(e^{n+1}) - 	an^{-1}(e^n)$.Now,the sum is a telescoping series:$S_N = \sum_{n=1}^N f(n) = (	an^{-1}(e^2) - 	an^{-1}(e)) + (	an^{-1}(e^3) - 	an^{-1}(e^2)) + \dots + (	an^{-1}(e^{N+1}) - 	an^{-1}(e^N))$.$S_N = 	an^{-1}(e^{N+1}) - 	an^{-1}(e)$.As $N 	o \infty$,$e^{N+1} 	o \infty$,so $	an^{-1}(e^{N+1}) 	o \frac{\pi}{2}$.Thus,$\sum_{n=1}^\infty f(n) = \frac{\pi}{2} - 	an^{-1}(e) = \cot^{-1}(e) = 	an^{-1}(\frac{1}{e})$.

If $f(n) = \tan^{-1} \left( \frac{e-1}{e^{-n} + e^{n+1}} \right)$ for all $n \in N$,then $\sum_{n=1}^\infty f(n)$ is equal to

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