The value of sumlimits_{n = 1}^infty {left( | vedclass

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Question

(A) Let the given sum be $S = \sum\limits_{n = 1}^\infty  {\left( {{{	an }^{ - 1}}\left( {\frac{n}{{n + 2}}} ight) - {{	an }^{ - 1}}\left( {\frac{

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$\frac{\pi }{4}$

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(A) Let the given sum be $S = \sum\limits_{n = 1}^\infty  {\left( {{{	an }^{ - 1}}\left( {\frac{n}{{n + 2}}} ight) - {{	an }^{ - 1}}\left( {\frac{{n - 1}}{{n + 1}}} ight)} ight)} $. This is a telescoping series of the form $\sum (f(n) - f(n-1))$. Let $f(n) = 	an^{-1}\left(\frac{n}{n+2}ight)$. The sum is $\lim_{N 	o \infty} \sum_{n=1}^{N} (f(n) - f(n-1))$. Expanding the sum: $S = (f(1) - f(0)) + (f(2) - f(1)) + (f(3) - f(2)) + \dots + (f(N) - f(N-1))$. $S = f(N) - f(0)$. $f(N) = 	an^{-1}\left(\frac{N}{N+2}ight) = 	an^{-1}\left(\frac{1}{1 + 2/N}ight)$. As $N 	o \infty$,$f(N) 	o 	an^{-1}(1) = \frac{\pi}{4}$. $f(0) = 	an^{-1}\left(\frac{0}{0+2}ight) = 	an^{-1}(0) = 0$. Therefore,$S = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

The value of $\sum\limits_{n = 1}^\infty {\left( {{{\tan }^{ - 1}}\left( {\frac{n}{{n + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{{n - 1}}{{n + 1}}} \right)} \right)} $ is equal to-

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