Question

Find the sum of infinite series :
${\cot }^{-1}\left({2.1}^2\right)+{\cot }^{-1}\left({2.2}^2\right)+{\cot }^{-1}\left({2.3}^2\right)+...+...$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is B $\frac{\pi }{4}$

Given, ${\cot }^{-1}2{\left(1\right)}^2+{\cot }^{2}2{\left(2\right)}^2+{\cot }^{-1}2{\left(3\right)}^2+{\cot }^{2}2{\left(4\right)}^2+...$
$\therefore T_n={\cot }^{-1}\left({2n}^2\right)={\tan }^{-1}\frac{1}{{2n}^2}$
$={\tan }^{-1}\left[\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right]={\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)$
$\therefore T_1={\tan }^{-1}3-{\tan }^{-1}1T_2={\tan }^{-1}5-{\tan }^{-1}3T_3={\tan }^{-1}7-{\tan }^{-1}5$
$\therefore S_n={\tan }^{-1}(2n+1)-{\tan }^{-1}1$
$\therefore \underset{n\to \infty }{lim}{S}_n={\tan }^{-1}\infty -\frac{\pi }{4}=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$