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)

$\displaystyle\sum_{r=1}^{\infty} T_r=\cot ^{-1}\left(r^2+\frac{3}{4}\right)=\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{1}{1+r^2-\frac{1}{4}}\right) $
$=\displaystyle\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) $
$=\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left(r+\frac{1}{2}\right)-\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left(r-\frac{1}{2}\right)$
Now it can be solved.