Maclaurin expansion $\log\left( \frac{1+x}{1-x}\right)$, show equality of two sums 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
$\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n}  = 2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$
?
 A: The expansion for $\log (1+x)$ is not  $\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}$, but
$$\begin{equation*}
\log (1+x)=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}%
x^{n+1}=\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n+1}}{n}x^{n+1}
\end{equation*}.$$
Consequently
$$
\begin{eqnarray*}
\log (1-x) &=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}\left(
-x\right) ^{n+1}=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}\left(
-1\right) ^{n+1}}{n+1}x^{n+1} \\
&=&-\sum_{n=0}^{\infty }\frac{1}{n+1}x^{n+1},
\end{eqnarray*}
$$
and
$$
\begin{eqnarray*}
\log \left( \frac{1+x}{1-x}\right)  &=&\log (1+x)-\log (1-x) \\
&=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}x^{n+1}+\sum_{n=0}^{
\infty }\frac{1}{n+1}x^{n+1} \\
&=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}+1}{n+1}x^{n+1} \\
&=&\sum_{n=0}^{\infty }\frac{2}{2n+1}x^{2n+1},
\end{eqnarray*}$$
because
$$
\begin{equation*}
\left( -1\right) ^{n}+1=\left\{ 
\begin{array}{c}
2\quad \text{if }n\text{ even} \\ 
0\quad \text{if }n\text{ odd}.
\end{array}
\right. 
\end{equation*}$$
A: It makes :
${\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^{n}}{n}}-(-1)^{n+1+n}\dfrac{x^{n}}{n}={\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^{n}}{n}}+\dfrac{x^{n}}{n}={\displaystyle \sum_{n=0}^{\infty}\dfrac{2x^{n}}{2n+1}}+{\displaystyle \sum_{n=1}^{\infty}-\dfrac{x^{n}}{2n}+\dfrac{x^{n}}{2n}}={\displaystyle \sum_{n=0}^{\infty}\dfrac{2x^{n}}{2n+1}}$ 
A: In fact you have $ (-1)^{n+1} (-x)^n = -(x)^n$. Thus if you look the term of the two sum you have: $x^n\frac{1 - (-1)^{n}}{n}$ and for the complete sum : $2\sum_{k .odd} \frac{x^k}{k}$ it's exactly that you want. 
