Question

The sum of the infinite terms of the series ${\cot }^{-1}(1^2+\frac{3}{4})+{\cot }^{-1}(2^2+\frac{3}{4})+.....$ is equal to

• A. ${\tan }^{-1}\left(1\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. ${\tan }^{-1}\left(3\right)$
• D. ${\tan }^{-1}\left(4\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(2\right)$

${\cot }^{-1}(1^2+\frac{3}{4})+{\cot }^{-1}(2^2+\frac{3}{4})+.....$
Let $t_n={\cot }^{-1}(n^2+\frac{3}{4})$
$={\tan }^{-1}(\frac{1}{n^2+\frac{3}{4}})$
$={\tan }^{-1}(\frac{1}{{1+(n}^2-\frac{1}{4})})$
$={\tan }^{-1}(\frac{1}{1+(n+\frac{1}{2})(n-\frac{1}{2})})$
$={\tan }^{-1}(\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n+\frac{1}{2})(n-\frac{1}{2})})$
$={\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}(n-\frac{1}{2})$
So, $S_n={\tan }^{-1}\left(\frac{3}{2}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{5}{2}\right)-{\tan }^{-1}\left(\frac{3}{2}\right)+......+{\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}(n-\frac{1}{2})$
$S_n={\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}\left(\frac{1}{2}\right)$
$S_{\infty }={lim}_{n\to \infty }{S}_n$
$={lim}_{n\to \infty }{\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}\left(\frac{1}{2}\right)$
$=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{2}\right)$
$={\cot }^{-1}\frac{1}{2}={\tan }^{-1}2$