Question

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

Answer 1

$\tan \left\{\underset{n\to \infty }{lim}\sum _{r=1}^n\left[{\tan }^{-1}\right(r+1)-{\tan }^{-1}r]\right\}(\because {\tan }^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)={\tan }^{-1}(r+1)-{\tan }^{-1}r)$
$=\tan \left(\underset{n\to \infty }{lim}{\tan }^{-1}\right(n+1)-\frac{\pi }{4})$
$=\tan \frac{\pi }{4}=1$