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

Answer: 1

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