Question:
$\lim _{n \rightarrow \infty} \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to
Solution:
$\lim _{n \rightarrow \alpha} \tan \left(\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r(r+1)}\right)\right)$
$=\lim _{n \rightarrow \alpha} \tan \left(\sum_{r=1}^{n} \tan ^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)\right)$
$=\tan \left(\lim _{n \rightarrow \alpha} \sum_{r=1}^{n}\left[\tan ^{-1}(r+1)-\tan ^{-1}(r)\right]\right)$
$=\tan \left(\lim _{n \rightarrow \infty}\left(\tan ^{-1}(n+1)-\frac{\pi}{4}\right)\right)$
$=\tan \left(\frac{\pi}{4}\right)=1$
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