sum from 1 to infinity (x^k)/(2k-1)

Question

$$\sum_{k=1}^\infty {\frac{x^k}{2k-1}} $$ Is there a closed form for this infinite sum? |x|< 1 .This is very similar to the infinite sum for log(1-z) but only being divided by odd integers. Can i modify the log function to arrive at this sum through Taylor expansion or is there some other method?

Answer 1

Let $x=y^2$ to make $$\sum_{k=1}^\infty {\frac{x^k}{2k-1}}=\sum_{k=1}^\infty {\frac{y^{2k}}{2k-1}}=\sum_{k=0}^\infty {\frac{y^{2k+2}}{2k+1}}=y\sum_{k=0}^\infty {\frac{y^{2k+1}}{2k+1}}=y\tanh ^{-1}(y)=\sqrt{x} \tanh ^{-1}\left(\sqrt{x}\right)$$