lim _{n rightarrow infty} sum_{r=1}^n cot ^{- | vedclass

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(C) We know that $\cot ^{-1}(x) = 	an ^{-1}\left(\frac{1}{x}ight)$.Thus,the given expression is $\lim _{n ightarrow \infty} \sum_{r=1}^n 	an ^{-

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$	an ^{-1} 2$

Vedclass · Answer

(C) We know that $\cot ^{-1}(x) = 	an ^{-1}\left(\frac{1}{x}ight)$.Thus,the given expression is $\lim _{n ightarrow \infty} \sum_{r=1}^n 	an ^{-1}\left(\frac{1}{r^2+\frac{3}{4}}ight)$.We can rewrite the argument as $\frac{1}{r^2+1-\frac{1}{4}} = \frac{1}{1+(r^2-\frac{1}{4})} = \frac{1}{1+(r-\frac{1}{2})(r+\frac{1}{2})}$.Using the identity $	an ^{-1}(a) - 	an ^{-1}(b) = 	an ^{-1}\left(\frac{a-b}{1+ab}ight)$,we have:$	an ^{-1}\left(\frac{(r+\frac{1}{2})-(r-\frac{1}{2})}{1+(r+\frac{1}{2})(r-\frac{1}{2})}ight) = 	an ^{-1}\left(r+\frac{1}{2}ight) - 	an ^{-1}\left(r-\frac{1}{2}ight)$.Now,the sum becomes a telescoping series:$S_n = \sum_{r=1}^n \left[	an ^{-1}\left(r+\frac{1}{2}ight) - 	an ^{-1}\left(r-\frac{1}{2}ight)ight]$$S_n = \left(	an ^{-1} \frac{3}{2} - 	an ^{-1} \frac{1}{2}ight) + \left(	an ^{-1} \frac{5}{2} - 	an ^{-1} \frac{3}{2}ight) + \dots + \left(	an ^{-1} \left(n+\frac{1}{2}ight) - 	an ^{-1} \left(n-\frac{1}{2}ight)ight)$.All intermediate terms cancel out,leaving $S_n = 	an ^{-1}\left(n+\frac{1}{2}ight) - 	an ^{-1}\left(\frac{1}{2}ight)$.Taking the limit as $n ightarrow \infty$:$\lim _{n ightarrow \infty} S_n = 	an ^{-1}(\infty) - 	an ^{-1}\left(\frac{1}{2}ight) = \frac{\pi}{2} - 	an ^{-1}\left(\frac{1}{2}ight) = \cot ^{-1}\left(\frac{1}{2}ight) = 	an ^{-1}(2)$.

$\lim _{n \rightarrow \infty} \sum_{r=1}^n \cot ^{-1}\left(r^2+\frac{3}{4}\right)=$

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