Question

The sum of the infinite series
${\left(\frac{1}{3}\right)}^2+{\frac{1}{3}\left(\frac{1}{3}\right)}^4+{\frac{1}{5}\left(\frac{1}{3}\right)}^6+$.... is

• A. $\frac{1}{4}{log}_{e}2$
• B. $\frac{1}{2}{log}_{e}2$
• C. $\frac{1}{6}{log}_{e}2$
• D. $\frac{1}{4}{log}_e\frac{3}{2}$

Answer 1

The correct option is C $\frac{1}{6}{log}_{e}2$

We know that,
$x+\frac{x^3}{3}+\frac{x^5}{5}+...$
$=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$
$\therefore {\left(\frac{1}{3}\right)}^2+\frac{1}{3}{\left(\frac{1}{3}\right)}^4+\frac{1}{5}{\left(\frac{1}{3}\right)}^6+...$
$=\frac{1}{3}[\left(\frac{1}{3}\right)+\frac{1}{3}{\left(\frac{1}{3}\right)}^3+\frac{1}{5}{\left(\frac{1}{3}\right)}^5+...]$
$=\frac{1}{3}.\frac{1}{2}\log \left(\frac{1+\frac{1}{3}}{1-\frac{1}{3}}\right)$
$=\frac{1}{6}\log \left(\frac{4}{2}\right)$
$=\frac{1}{6}\log 2$