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

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Question

(A) We know that $r^4+r^2+1 = (r^2-r+1)(r^2+r+1)$.Also,$r^2+r+1 - (r^2-r+1) = 2r$.Therefore,the term inside the summation can be written as:$	an ^{-1

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) We know that $r^4+r^2+1 = (r^2-r+1)(r^2+r+1)$.Also,$r^2+r+1 - (r^2-r+1) = 2r$.Therefore,the term inside the summation can be written as:$	an ^{-1}\left(\frac{2r}{1+(r^4+r^2+1)}ight) = 	an ^{-1}\left(\frac{(r^2+r+1)-(r^2-r+1)}{1+(r^2+r+1)(r^2-r+1)}ight) = 	an ^{-1}(r^2+r+1) - 	an ^{-1}(r^2-r+1)$.Now,the sum is a telescoping series:$S_n = \sum_{r=1}^n \{	an ^{-1}(r^2+r+1) - 	an ^{-1}(r^2-r+1)\}$$S_n = (	an ^{-1}(3) - 	an ^{-1}(1)) + (	an ^{-1}(7) - 	an ^{-1}(3)) + \dots + (	an ^{-1}(n^2+n+1) - 	an ^{-1}(n^2-n+1))$$S_n = 	an ^{-1}(n^2+n+1) - 	an ^{-1}(1)$.Taking the limit as $n ightarrow \infty$:$\lim _{n ightarrow \infty} S_n = \lim _{n ightarrow \infty} (	an ^{-1}(n^2+n+1) - \frac{\pi}{4}) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

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

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