I want to show $$\sum^{\infty}_{k=1} \frac{x^{2k-1}}{2k-1} \equiv \frac{1}{2}\left[\sum^{\infty}_{n=1}\frac{x^{n}}{n} - \sum^{\infty}_{n=1}\frac{(-x)^{n}}{n}\right]$$
Although I'm really not sure where to start, can anyone give me an idea of a direction to start on?
Note that $$(-x)^n=(-1)^nx^n=\begin{cases}\phantom{-1\cdot!}x^n, & n=2k \\ -1\cdot x^n, & n=2k-1\end{cases}, k \in \mathbb N$$Hence, starting from the RHS (which is more complicated and therefore a better idea) you have that $$\begin{align*}\left[\sum^{\infty}_{n=1}\frac{x^{n}}{n} - \sum^{\infty}_{n=1}\frac{(-x)^{n}}{n}\right]&= \sum^{\infty}_{n=1}\frac{x^n-(-x)^{n}}{n}=\\&=\sum^{\infty}_{n=2k}\frac{x^{n}-x^{n}}{n}+\sum^{\infty}_{n=2k-1}\frac{x^{n}+x^{n}}{n}=\\&=2\sum^{\infty}_{n=1}\frac{x^{2k-1}}{2k-1}\end{align*}$$ Divide with $2$ to conclude.