<s>[INST] 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!
 [/INST] Let $$f(x):=\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1}.$$ The radius of convergence is $1$.
Then
$$f'(x):=\sum_{k=0}^\infty{x^{2k}}=\sum_{k=0}^\infty{(x^2)^k}=\frac1{1-x^2}.$$
By integration,
$$f(x)=\int_0^x\frac{dx}{1-x^2}=\frac12\int_0^x\left(\frac1{1+x}+\frac1{1-x}\right)dx=\frac12\log\left|\frac{1+x}{1-x}\right|$$ which you evaluate at $x=\dfrac15$.
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