# 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}

• Hem, is this a positive number ??
– user65203
Commented Sep 21, 2018 at 8:20

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 $$x=\dfrac15$$.

• thanks u @Yvves,,,,,,this one i like its Commented Sep 21, 2018 at 8:37

$$\frac15+\frac13\frac1{5^3}+\frac15\frac1{5^5}+\cdots=\frac12\log\frac{1+\dfrac15}{1-\dfrac15}=\frac12\log\frac32.$$
The series is fast converging. With the first five terms, $$0.202732552127\cdots$$, vs. the exact value $$0.2027325540541\cdots$$.