Unsure how to simplify power series If I want to find the power series representation of the following function:
$$ \ln \frac{1+x}{1-x} $$
I understand that it can be written as 
$$ \ln (1+x) - \ln(1-x) $$
And I understand that if I now write in the power series representations for $ln(1+x)$ and $ln(1-x)$:
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n} $$
My textbook solution does an odd thing where it writes it out as
$$\sum_{n=1}^\infty \frac{x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n} $$
$$2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}  $$
I have no idea how it got from the line where I have the power series representation for $ln(1+x)$ and $ln(1-x)$ to the last two lines. If anyone could help me link my part to the textbook solution I would really appreciate it! Thank you! 
 A: Assuming $|x|<1$, by absolute convergence, one is allowed to rearrange terms, obtaining
$$
\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n}=\sum_{n=1}^\infty \frac{(-1)^{n-1}+1}{n} \cdot x^{n}=2\sum_{k=1}^\infty \frac{x^{2k-1}}{2k-1}
$$ since term corresponding to $n=2,4,\cdots,$ cancel and since $(-1)^{n-1}+1=2$ for $n=1,3,5\cdots$.
A: $$\begin{align}\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n}&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-1)^nx^{n}}{n}\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{2n-1}x^{n}}{n}\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{-x^{n}}{n}\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} + \sum_{n=1}^\infty \frac{x^{n}}{n}\\
&=\left[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\right]+ \left[x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\dots\right]\\
&=2\left[\frac{x^1}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\dots\right]\\
&=2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}
\end{align}$$
