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: $$\frac{1}{2}\ln1.5.$$