Question

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

• A. $\frac{1}{4}\ln \left(2\right)$
• B. $\frac{1}{2}\ln \left(2\right)$
• C. $\frac{1}{6}\ln \left(2\right)$
• D. $\frac{1}{8}\ln \left(2\right)$

Answer 1

The correct option is C

$\frac{1}{6}\ln \left(2\right)$

Find the sum of the given series

${\left(\frac{1}{3}\right)}^2+\left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^4+\left(\frac{1}{5}\right){\left(\frac{1}{3}\right)}^6+...$

$$\begin{gathered} =\left(\frac{1}{3}\right)\times \left\{{\left(\frac{1}{3}\right)}^1+\left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^3+\left(\frac{1}{5}\right){\left(\frac{1}{3}\right)}^5+...\right\} \\ =\left(\frac{1}{3}\right)\times \left(\frac{1}{2}\right)\ln \left(\frac{\left(1+{\displaystyle \frac{1}{3}}\right)}{\left(1-{\displaystyle \frac{1}{3}}\right)}\right) \\ =\frac{1}{6}\times \ln \left(2\right) \end{gathered}$$ ;$\left[\ln \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+......\infty \right)\right]$

Hence, the correct answer is $C$.