Question

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

• A. ${\tan }^{-1}1$
• B. ${\tan }^{-1}2$
• C. $-{\tan }^{-1}1$
• D. $\pi -{\tan }^{-1}2$

Answer 1

The correct option is B ${\tan }^{-1}2$

Let $S_n={\cot }^{-1}(1^2+\frac{3}{4})+{\cot }^{-1}(2^2+\frac{3}{4})+{\cot }^{-1}(3^2+\frac{3}{4})+\cdots$
$T_n={\cot }^{-1}(n^2+\frac{3}{4})={\tan }^{-1}\left(\frac{1}{n^2+\frac{3}{4}}\right)$
$={\tan }^{-1}\left(\frac{1}{1+(n^2-\frac{1}{4})}\right)$
$\Rightarrow T_n={\tan }^{-1}\left(\frac{1}{1+(n-\frac{1}{2})(n+\frac{1}{2})}\right)$

$\Rightarrow T_n={\tan }^{-1}\left(\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n-\frac{1}{2})(n+\frac{1}{2})}\right)$

$\Rightarrow T_n={\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}(n-\frac{1}{2})$

Now, $S_n=\sum _{n=1}^{\infty }{T}_n=\sum _{n=1}^{\infty }\{{\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}(n-\frac{1}{2})\}$
$={\tan }^{-1}\left(\frac{3}{2}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{5}{2}\right)-{\tan }^{-1}\left(\frac{3}{2}\right)+{\tan }^{-1}\left(\frac{7}{2}\right)-{\tan }^{-1}\left(\frac{5}{2}\right)\cdots +{\tan }^{-1}\left(\infty \right)$
$={\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(\frac{1}{2}\right)$
$=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{2}\right)$
$={\cot }^{-1}\left(\frac{1}{2}\right)={\tan }^{-1}\left(2\right)$