Find the sum of infinite series

Question

Find the sum of infinite series $$\frac{1}{5}+\frac{1}{3}.\frac{1}{5^3}+\frac{1}{5}.\frac{1}{5^5}+...$$

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I'm trying by consider this sum as S and then multiply $\frac{1}{5}$ and substract from S but i can't found any sum....

Answer 1

Let $\frac{1}{5}=x$ and $f(x)=x+\frac{1}{3}x^3+\frac{1}{5}x^5+...$

Thus, $$f'(x)=1+x^2+x^4+...=\frac{1}{1-x^2}$$ and $$f(x)=\int\limits_0^x\frac{1}{1-t^2}dt=\ln\sqrt{\frac{1+x}{1-x}},$$ which after substitution $x=\frac{1}{5}$ gives the answer: $$\frac{1}{2}\ln1.5.$$

Answer 2

$$(\tanh^{-1}(x))'=1+x^2+x^4\cdots=\frac{1}{1-x^2}= \frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x} \right)$$

hence for $$\tanh^{-1}(x)=x+\frac{x^3}{3} +\frac{x^5}{5}+\frac{x^7}{7}\cdots= \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$$

taking $x=\frac{1}{5}$ you have $$ \tanh^{-1}\left(\frac{1}{5}\right) = \frac{1}{5}+\frac{1}{3}.\frac{1}{5^3}+\frac{1}{5}.\frac{1}{5^5}+...= \frac{1}{2}\ln\left(\frac{3}{2}\right)$$