Q. The sum of the infinite series $\left( \frac{1}{3} \right)^{2} + \frac{1}{3}\left( \frac{1}{3} \right)^{4} + \frac{1}{5}\left( \frac{1}{3} \right)^{6} + \ldots$ is
A) $\frac{1}{4}\log_{e}2$
B) $\frac{1}{2}\log_{e}2$
C) $\frac{1}{6}\log_{e}2$
D) $\frac{1}{4}\log_{e}\frac{3}{2}$
Explanation
(c)
$\left( \frac{1}{3} \right)^{2} + \frac{1}{3}\left( \frac{1}{3} \right)^{4} + \frac{1}{5}\left( \frac{1}{3} \right)^{6} + ...$
$= \frac{1}{3}\left\lbrack \left( \frac{1}{3} \right) + \frac{1}{3}\left( \frac{1}{3} \right)^{3} + \frac{1}{5}\left( \frac{1}{3} \right)^{5} + ... \right\rbrack$
$= \frac{1}{3}.\frac{1}{2}\log\left( \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}} \right)\ \ \ \ \ \left\lbrack \because\frac{1}{2}\log{\left( \frac{1 + x}{1 - x} \right) = x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ...\ } \right\rbrack$
$= \frac{1}{6}\log_{e}2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.