Question

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

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

Answer 1

Answer: (B), (D)

$\sum _{r=1}^{\infty }{T}_r={\cot }^{-1}\left(r^2+\frac{3}{4}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{1+r^2-\frac{1}{4}}\right)$
$=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r-\frac{1}{2}\right)\left(r+\frac{1}{2}\right)}\right)$
$=\sum _{r=1}^{\infty }{\tan }^{-1}\left(r+\frac{1}{2}\right)-\sum _{r=1}^{\infty }{\tan }^{-1}\left(r-\frac{1}{2}\right)$
Now it can be solved.