Question

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

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

Answer 1

Answer: (C)

$\displaystyle \sum_{r=2}^{\infty} \tan ^{-1}\left(\frac{(r-2)-(r-3)}{1+(r-3)(r-2)}\right)=\displaystyle\sum_{r=2}^{\infty}\left(\tan ^{-1}(r-2)-\tan ^{-1}(r-3)\right)$
$= \tan ^{-1} 0-\tan ^{-1}(-1) $
$ \tan ^{-1} 1-\tan ^{-1} 0 $
$\tan ^{-1} 2-\tan ^{-1} 1 $
$\vdots $
$ \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} $