Calculating the sum of the infinite series $\frac{1}{5} + \frac{1}{3}\frac{1}{5^3} + \frac{1}{5} \frac{1}{5^5} +......$ How do I calculate the sum of the infinite  series?
$$\frac{1}{5} + \frac{1}{3}.\frac{1}{5^3} + \frac{1}{5}. \frac{1}{5^5} +......$$
My attempt  :
I know  that $$\log (\frac{1+x}{1-x}) = 2 \, \left(x + \frac{x^3}{3}  + \frac{x^5}{5} +.....+\frac{x^{2r-1}} {2r-1}+..\right)$$
Now
\begin{align}\frac{1}{5} + (\frac{1}{3}.\frac{1}{5^3} + \frac{1}{5}. \frac{1}{5^5} +......) &=  \frac{1}{5} + \log \frac{(1 +1/5)} {(1-1/5)} - \frac{5}{2} \\ &=\frac{1}{5} + \log (\frac{6}{4})-10= -\frac{49}{5}  + \log \frac{3}{2} 
\end{align}
Please verify my answer, thank you!

From your own formula,
$$\frac15+\frac13\frac1{5^3}+\frac15\frac1{5^5}+\cdots=\frac12\log\frac{1+\dfrac15}{1-\dfrac15}=\frac12\log\frac32.$$

Check:
The series is fast converging. With the first five terms, $0.202732552127\cdots$, vs. the exact value $0.2027325540541\cdots$.
