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

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(A) We know that $	an ^{-1}\left(\frac{1}{1+r+r^{2}}ight) = 	an ^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}ight) = 	an ^{-1}(r+1) - 	an ^{-1}(r)$.Thus

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(A) We know that $	an ^{-1}\left(\frac{1}{1+r+r^{2}}ight) = 	an ^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}ight) = 	an ^{-1}(r+1) - 	an ^{-1}(r)$.Thus,the sum is a telescoping series:$\sum_{r=1}^{n} \left[	an ^{-1}(r+1) - 	an ^{-1}(r)ight] = (	an ^{-1}(2) - 	an ^{-1}(1)) + (	an ^{-1}(3) - 	an ^{-1}(2)) + \dots + (	an ^{-1}(n+1) - 	an ^{-1}(n))$.This simplifies to $	an ^{-1}(n+1) - 	an ^{-1}(1) = 	an ^{-1}(n+1) - \frac{\pi}{4}$.Now,taking the limit as $n ightarrow \infty$:$\lim _{n}$ ${ightarrow \infty} 	an \left(	an ^{-1}(n+1) - \frac{\pi}{4}ight) = 	an \left(\frac{\pi}{2} - \frac{\pi}{4}ight) = 	an \left(\frac{\pi}{4}ight) = 1$.

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

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