Question

$\underset{n\to \infty }{\lim }\sum _{r=1}^n{\tan }^{-1}\left(\frac{2r}{r^4+r^2+2}\right)=$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{-\pi }{4}$
• D. $\frac{-\pi }{2}$

Answer 1

Answer: (A)

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