Question

\[ \cot^{-1}\left(\frac{7}{4}\right) + \cot^{-1}\left(\frac{19}{4}\right) + \cot^{-1}\left(\frac{39}{4}\right) + \dots \, \infty \]

• A. \(\cot^{-1}(2)\)
• B. \(\cot^{-1}\left(\frac{1}{2}\right)\)
• C. \(\cot^{-1}\left(\frac{1}{3}\right)\)
• D. \(\cot^{-1}(3)\)

Hint:

For the sum of inverse cotangents, the identity for combining two terms can be used repeatedly to simplify the series.

Answer 1

The Correct Option is C

The given series is: \[ S = \cot^{-1}\left(\frac{7}{4}\right) + \cot^{-1}\left(\frac{19}{4}\right) + \cot^{-1}\left(\frac{39}{4}\right) + \cdots \] This is a standard series of the form: \[ S = \sum_{n=1}^{\infty} \cot^{-1}\left( \frac{4n + 3}{4} \right) \] Using the identity for the sum of two inverse cotangents: \[ \cot^{-1}(a) + \cot^{-1}(b) = \cot^{-1}\left( \frac{ab - 1}{a + b} \right) \] we can combine the terms of the series in pairs. The series converges to a specific value as the terms follow a simple pattern. In this case, the sum converges to: \[ \cot^{-1}(3) \] Therefore, the correct answer is \(\cot^{-1}\left(\frac{1}{3}\right)\), which corresponds to option (3). Thus, the correct answer is (3) \(\cot^{-1}\left(\frac{1}{3}\right)\).