Question

The value of $\sum _{r=2}^{\infty }{\tan }^{-1}\left(\frac{1}{r^2-5r+7}\right)$, is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. $\frac{5\pi }{4}$

Answer 1

Answer: (C)

$\sum _{r=2}^{\infty }{\tan }^{-1}\left(\frac{(r-2)-(r-3)}{1+(r-3)(r-2)}\right)=\sum _{r=2}^{\infty }\left({\tan }^{-1}(r-2)-{\tan }^{-1}(r-3)\right)$
$={\tan }^{-1}0-{\tan }^{-1}(-1)$
${\tan }^{-1}1-{\tan }^{-1}0$
${\tan }^{-1}2-{\tan }^{-1}1$
$⋮$
${\tan }^{-1}(n-2)-{\tan }^{-1}(n-3)$
$S_n={\tan }^{-1}(n-2)+\frac{\pi }{4}$
$\therefore S_{\infty }=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$