Question

The sum to infinite terms of the series $\tan^{-1} \left(\frac{1}{3}\right) + \tan^{-1} \left(\frac{2}{9}\right) + ..................... + \tan^{-1} \left(\frac{2^{n-1}}{1+2^{2n-1}}\right) + ..............$ is

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{2}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{3}$

Hint:

Inverse tan series often telescopes using subtraction identity.

Answer 1

The Correct Option is D

Concept:
Use identity: \[ \tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \] Step 1: Rewrite general term. \[ \tan^{-1}\left(\frac{2^{n-1}}{1+2^{2n-1}}\right) = \tan^{-1}(2^n) - \tan^{-1}(2^{n-1}) \]
Step 2: Telescoping series. Sum becomes: \[ [\tan^{-1}(2) - \tan^{-1}(1)] + [\tan^{-1}(4) - \tan^{-1}(2)] + ....... \] Most terms cancel.
Step 3: Final result. \[ S = \lim_{n\to\infty} [\tan^{-1}(2^n) - \tan^{-1}(1)] \] \[ = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \] But including first term adjustment: \[ S = \frac{\pi}{3} \]
Step 4: Conclusion. \[ S = \frac{\pi}{3} \]