Question

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

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

Answer 1

The correct option is B $\frac{\pi }{2}$

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

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

$=limn\to \infty {\sum }_{r=1}^n{\tan }^{-1}\left(\frac{r(r+1)-r(r-1)}{1+\left[r\right(r-1\left)\right]\left[r\right(r+1\left)\right]}\right)$

$=limn\to \infty {\sum }_{r=1}^n{\tan }^{-1}\left(r\right(r+1)-{\tan }^{-1}\left(r\right(r-1\left)\right)$

$=limn\to \infty \left[ta{n}^{-1}\right(1(1+1)-{\tan }^{-1}\left(1\right(1-1\left)\right)+ta{n}^{-1}\left(2\right(2+1\left)\right)-{\tan }^{-1}\left(2\right(2-1\left)\right)+.......+{\tan }^{-1}\left(n\right(n+1\left)\right)-{\tan }^{-1}\left(n\right(n-1\left)\right)]$

$=limn\to \infty \left[ta{n}^{-1}\right(2)-{\tan }^{-1}(0)+ta{n}^{-1}(6\left)\right)-{\tan }^{-1}\left(2\right)+.........+{\tan }^{-1}\left(n\right(n+1\left)\right)-{\tan }^{-1}\left(n\right(n-1\left)\right)]$

$=limn\to \infty [-{\tan }^{-1}(0)+{\tan }^{-1}(n(n+1)\left)\right]$

$=limn\to \infty \left[{\tan }^{-1}\right(n(n+1)\left)\right]$

$=limn\to \infty \left[{\tan }^{-1}\right(n^2+n\left)\right)]$

$={\tan }^{-1}\left(\infty \right)$

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