TITLE: Taylor series expansion of $(x^2-x)\ln(1-x)$ and calculating a sum
QUESTION [1 upvotes]: So I have this function: $$f(x)=(x^2-x)\ln(1-x)$$
So I want to calculate it's Taylor series centered at x=0, basically that is Maclaurin series, and that series will be of help when calculating this sum:
$$\sum_{n=1}^{\infty} \frac{1}{2^n n (n+1)}$$
So it's obvious that the sum is going to be converging, as far as I know.
I am trying to calculate the sum, but the Taylro series is giving me problems, I know how to calculate for $f(x)=ln(1+x)$, since it's a known Taylor series, but i don't knwo how to apply it here.
Any help would be appreciated.

REPLY [5 votes]: By the Taylor series expansion,$$
\log (1-x)=-\sum_{n=1}^{\infty} \frac{x^n}n, \quad |x|<1, \tag1
$$ one may write, for $|x|<1$,
$$
\begin{align}
(x^2-x)\log (1-x)&=(x-x^2)\sum_{n=1}^{\infty} \frac{x^n}n
\\&=\sum_{n=1}^{\infty} \frac{x^{n+1}}n-\sum_{n=1}^{\infty} \frac{x^{n+2}}n
\\&=\sum_{n=2}^{\infty} \frac{x^{n}}{n-1}-\sum_{n=3}^{\infty} \frac{x^{n}}{n-2}
\\&=x^2-\sum_{n=3}^{\infty} \frac{x^{n}}{(n-1)(n-2)}
\end{align}
$$ from which, by putting $x=\dfrac12$, one deduces 
$$
\sum_{n=1}^{\infty} \frac{1}{2^n n (n+1)}=1-\ln 2.
$$
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