Question

The sum of the infinite series $ \cot^{-1} \left( \frac{7}{4} \right) + \cot^{-1} \left( \frac{19}{4} \right) + \cot^{-1} \left( \frac{39}{4} \right) + \cot^{-1} \left( \frac{67}{4} \right) + ... $ is:

• A. \( \frac{\pi}{2} + \tan^{-1} \left( \frac{1}{2} \right) \)
• B. \( \frac{\pi}{2} - \cot^{-1} \left( \frac{1}{2} \right) \)
• C. \( \frac{\pi}{2} + \cot^{-1} \left( \frac{1}{2} \right) \)
• D. \( \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{2} \right) \)

Hint:

When dealing with infinite series involving cotangent or tangent, look for cancellation patterns and simplify the terms.

Answer 1

The Correct Option is D

Let \( S \) represent the sum of the series, with the general term given by \( T_n \):\[T_n = \cot^{-1} \left( \frac{4n}{2n^2 + 3} \right)\]Simplifying \( T_n \):\[T_n = \cot^{-1} \left( \frac{n + \frac{1}{2}}{1 + \left( n + \frac{1}{2} \right)^2} \right)\]The series is then expressed as:\[S = T_1 + T_2 + \dots = \cot^{-1} \left( n + \frac{1}{2} \right) - \cot^{-1} \left( n - \frac{1}{2} \right)\]
The sum of the infinite series is:\[S = \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{2} \right)\]
Therefore, the definitive answer is \( \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{2} \right) \).