<s>[INST] 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?
 [/INST] The derivative is
$$\biggl(\ln\frac{1+x}{1-x}\biggr)'=\frac{1-x}{1+x}\,\frac 2{(1-x)^2}=\frac 2{1-x^2}$$
Now
$$\frac 2{1-x^2}=2\sum_{n=0}^\infty x^{2n},\enspace\text{so}\quad\ln\frac{1+x}{1-x}=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}, $$
taking into account that both sides are $0$ for $x=0$.
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