Question

$\cot^{-1}(2 \cdot 1^2) + \cot^{-1}(2 \cdot 2^2) + \cot^{-1}(2 \cdot 3^2) + \dots \dots \infty =$

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{8}$

Hint:

To sum $\tan^{-1}$ series, express the argument as $\frac{x-y}{1+xy}$ to create a telescoping sum.

Answer 1

The Correct Option is C

Step 1: General Term
$T_r = \cot^{-1}(2r^2) = \tan^{-1}\left(\frac{1}{2r^2}\right) = \tan^{-1}\left(\frac{2}{4r^2}\right)$.

Step 2: Method of Differences
$T_r = \tan^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right) = \tan^{-1}(2r+1) - \tan^{-1}(2r-1)$.

Step 3: Summation
$\sum_{r=1}^{n} T_r = [\tan^{-1}(3)-\tan^{-1}(1)] + [\tan^{-1}(5)-\tan^{-1}(3)] + \dots + [\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]$.
Sum $= \tan^{-1}(2n+1) - \tan^{-1}(1)$.

Step 4: Evaluate for $\infty$
As $n \to \infty$, Sum $= \tan^{-1}(\infty) - \tan^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.
Final Answer: (C)