Let S_{n} = cot^{-1} 2 + cot^{-1} 8 + cot^{-1 | vedclass

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(B) The $n^{	ext{th}}$ term of the series is $t_{n} = \cot^{-1}(2n^2)$.Using the identity $\cot^{-1} x = 	an^{-1} \frac{1}{x}$, we have $t_{n} = 	a

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$\frac{\pi}{4}$

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(B) The $n^{	ext{th}}$ term of the series is $t_{n} = \cot^{-1}(2n^2)$.Using the identity $\cot^{-1} x = 	an^{-1} \frac{1}{x}$, we have $t_{n} = 	an^{-1} \frac{1}{2n^2}$.We can rewrite this as $t_{n} = 	an^{-1} \frac{2}{4n^2} = 	an^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)}$.Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \frac{x-y}{1+xy}$, we get $t_{n} = 	an^{-1}(2n+1) - 	an^{-1}(2n-1)$.The sum $S_{n} = \sum_{k=1}^{n} t_{k} = (	an^{-1} 3 - 	an^{-1} 1) + (	an^{-1} 5 - 	an^{-1} 3) + \dots + (	an^{-1}(2n+1) - 	an^{-1}(2n-1))$.This is a telescoping series, so $S_{n} = 	an^{-1}(2n+1) - 	an^{-1} 1$.Taking the limit as $n ightarrow \infty$, $\lim_{n ightarrow \infty} S_{n} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

Let $S_{n} = \cot^{-1} 2 + \cot^{-1} 8 + \cot^{-1} 18 + \cot^{-1} 32 + \dots$ to $n^{\text{th}}$ term. Then $\lim_{n \rightarrow \infty} S_{n}$ is

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