Taylor series expansion of $(x^2-x)\ln(1-x)$ and calculating a sum

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.

• The Taylor series of $(x^2-x)ln(1-x)$ is simply the Taylor series of $ln(1-x)$ multiplied by $(x^2-x)$. Commented Aug 4, 2016 at 7:51
• why not integrating $\sum_{n\geq 0}x^{n-1}$ two times? Commented Aug 4, 2016 at 7:54
• $ln(1-x)=ln(1+(-x))$ Commented Aug 4, 2016 at 8:19

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 $$\sum_{n=1}^{\infty} \frac{1}{2^n n (n+1)}=1-\ln 2.$$