My Problem is to expand $f(x)=\dfrac{\ln(1-x)}{1+x}$ into a power series.
My Approach: from looking onto the Graphs of this function, i know, for rising x the y is falling towards Zero, without reaching it. but i don't think there is convergence for the series.
A power series has the scheme: $\sum\limits_{n=0}^{\infty} a_{n}\cdot x^{n}$ but im stuck in trying to convert $f(x)$ into a suitable sequence $a_{n}$.
Any hints?
You have two series expansions, both for $|x|<1$, namely:
1) $\log(1+x)=\displaystyle\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{x^n}{n}$
2) $\frac{1}{1-x}=\displaystyle\sum_{n=0}^{+\infty}x^n$
Substituting $x$ with $-x$ doesn't change the convergence radius, hence you get
1') $\log(1-x)=-\displaystyle\sum_{n=1}^{+\infty}\frac{x^n}{n}$
2') $\frac{1}{1+x}=\displaystyle\sum_{n=0}^{+\infty}(-1)^nx^n$
Now multiply the two series together
$$\frac{\log(1-x)}{1+x}=(1-x+x^2-x^3+\ldots)\cdot(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldots)$$ $$=-x+\frac{x^2}{2}-\frac{5x^3}{6}+\frac{7x^4}{12}-\frac{47x^5}{60}+ O(x^6)$$
