<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] Beginning with 
$$\frac{1}{1 - x} = \sum_{k = 0}^{\infty} x^{k}, $$
we can integrate to get $$\log(1 - x) = -\sum_{k = 1}^{\infty} \frac{x^{k}}{k}.$$
Also, if we plug in $(-x)$ for $x$ into the equation for $\frac{1}{1 - x}$, we get 
$$\frac{1}{1 + x} = \sum_{k = 0}^{\infty}(-x)^{k}.$$
Integrating, we get
$$\log(1 + x) = \sum_{k = 1}^{\infty} \frac{(-x)^{k}}{k}.$$
Now, by properties of $\log$, $\frac{\log(1 + x)}{\log(1 - x)} = \log(1 + x) - \log(1 - x)$. So,
$$\frac{\log(1 + x)}{\log(1 - x)} = \sum_{k = 1}^{\infty} \frac{(-x)^{k}}{k} + \sum_{k = 1}^{\infty} \frac{x^{k}}{k}$$
$$= \sum_{k = 1}^{\infty} \frac{(-x)^{k} + x^{k}}{k}.$$
Note that when $k$ is odd, the terms vanish.
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