Question

The sum $\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{3}{n^2+n-1}\right)$ is equal to

• A. $-\frac{3\pi }{4}+{\cot }^{-1}2$
• B. $\frac{3\pi }{4}+{\cot }^{-1}2$
• C. $\frac{\pi }{2}+{\tan }^{-1}2$
• D. $\frac{\pi }{2}+{\cot }^{-1}3$

Answer 1

The correct option is B $\frac{3\pi }{4}+{\cot }^{-1}2$

Let $S_{\infty }=\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{3}{n^2+n-1}\right)$
$=\sum _{n=1}^{\infty }{\tan }^{-1}\left[\frac{(n+2)-(n-1)}{1+(n-1)(n+2)}\right]$
$=\sum _{n=1}^{\infty }{\tan }^{-1}(n+2)-{\tan }^{-1}(n-1)$
$\therefore T_1={\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(0\right)$
$T_2={\tan }^{-1}\left(4\right)-{\tan }^{-1}\left(1\right)$
$T_3={\tan }^{-1}\left(5\right)-{\tan }^{-1}\left(2\right)$
$T_4={\tan }^{-1}\left(6\right)-{\tan }^{-1}\left(3\right)$
$\begin{array}{ccc}⋮& ⋮& ⋮\end{array}$
$T_n={\tan }^{-1}(n+2)-{\tan }^{-1}(n-1)$
Adding all the terms, we get
$S_n={\tan }^{-1}(n+2)+{\tan }^{-1}(n+1)+{\tan }^{-1}\left(n\right)-{\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(2\right)$
$\therefore S_{\infty }=\frac{\pi }{2}+\frac{\pi }{2}+\frac{\pi }{2}-\frac{\pi }{4}-(\frac{\pi }{2}-{\cot }^{-1}2)$
$=\frac{3\pi }{4}+{\cot }^{-1}2$