I am supposed to find the Maclaurin expantion of
$ \log\left( \frac{1+x}{1-x} \right) $
So I noticed the obvious that $\log(1+x) - \log(1-x)$
Then Maclaurin polynomial of $\log (1+x)$ equals $\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}$
So by doing a quick quick change of variables (-x) I obtained the maclaurin expansion of $\log(1-x)$
In total, I obtained that the Maclaurin expansion of $ \log\left( \frac{1+x}{1-x} \right) $ equals.
$ \displaystyle P(x) = \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n}$
But in my book the answer says $2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$, I can see that this is correct, by writing out a few terms. But how do I show this by algebra?
My question is therefore, how do I show that
$\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n} = 2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$
?