Taylor series of $\ln\frac{1+x}{1-x}$ 
Let $f(x)=\ln\frac{1+x}{1-x}$ for $x$ in $(-1,1)$. Calculate the Taylor series of $f$ at $x_0=0$

I determined some derivatives:
$f'(x)=\frac{2}{1-x^2}$;  $f''(x)=\frac{4x}{(1-x^2)^2}$;    $f^{(3)}(x)=\frac{4(3x^2+1)}{(1-x^2)^3}$;  $f^{(4)}(x)=\frac{48x(x^2+1)}{(1-x^2)^4}$; $f^{(5)}(x)=\frac{48(5x^2+10x^2+1)}{(1-x^2)^5}$
and their values at $x_0=0$:
$f(0)=0$; $f'(0)=2$; $f''(0)=0$; $f^{(3)}(0)=4=2^2$
$f^{(4)}(0)=0$; $f^{(5)}(0)=48=2^4.3$; $f^{(7)}(0)=1440=2^5.3^2.5$
I can just see that for $n$ even, $f^{(n)}(0)=0$, but how can I generalize the entire series?

\begin{align}
\ln \frac{1+x}{1-x} &= \ln (1+x) - \ln (1-x) \\
&= \sum_{n=0}^\infty \frac{(-1)^n}{n+1}x^{n+1} - \sum_{n=0}^\infty \frac{(-1)^n}{n+1}(-x)^{n+1}\\
&= \sum_{n=0}^\infty \frac{(-1)^n}{n+1}x^{n+1} + \sum_{n=0}^\infty \frac{1}{n+1}x^{n+1}\\
&= \sum_{n=0}^\infty \frac{(-1)^n+1}{n+1}x^{n+1}\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}+1}{n}x^n\\
&=\sum_{n=1}^\infty \frac{2}{2n-1}x^{2n-1}\\
\end{align}
Remark: Your observation that all the even terms vanishes is due to this is an odd function. 
