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

AP EAMCETAP EAMCET 2018

Solution:

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