Question

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

• A. $\pi /4$
• B. $\pi /2$
• C. $\frac{3\pi }{4}$
• D. None of these

Answer 1

The correct option is B $\pi /2$

Given,

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

NOw,

$ta{n}^{-}1\left(\frac{(r^2+r)-(r^2-r)}{1+(r^2+r)(r^2-r)}\right)=ta{n}^{-}1(r^2+r)-ta{n}^{-}1(r^2-r)$

summing over $r=1$ to $r=n$,the terms will get cancel and we get

$=>ta{n}^{-}1(n^2+n)-ta{n}^{-}1\left(0\right)$

$=>{lim}_{n\to \infty }ta{n}^{-}1(n^2+n)$

$=>ta{n}^{-}1\left(\infty \right)$

$=>\frac{\pi }{2}$