When finding the taylor series for $\frac{1}{(x-1)^3}$, work is given as $\frac{1}{(x-1)^3}=-\left(\frac{1}{2}\right) \frac{d^2}{d x^2} \frac{1}{1-x}=-\sum_{n=0}^{\infty}\left(\frac{(n+2)(n+1) x^n}{2}\right)$. I do not understand $-\left(\frac{1}{2}\right) \frac{d^2}{d x^2} \frac{1}{1-x}=-\sum_{n=0}^{\infty}\left(\frac{(n+2)(n+1) x^n}{2}\right)$: as I would integrate $\frac 1 {1-x} = \sum _{n=0}^\infty x^n $ twice , getting $\sum _{n=0}^\infty \frac {x^{n+2}}{(n+1)(n+2)}$, and multiply this by $-1 \over 2$. I would appreciate an explanation as to this part (I understand why this taylor series is correct using other means).
UPDATE: thanks to the comments, I have seen my mistake in integrating twice rather than differentiating twice. However, I now have a new question about this. When I differentiate twice, I get $\sum_ {n=0}^\infty n(n-1)x^{n-2}$. Why do we "shift over" the indices to get $\sum_{n=0}^{\infty}{(n+2)(n+1) x^n})$?
Note that $$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$ Thus $$\frac{d}{dx}\frac{1}{1-x}=\frac{d}{dx}\sum_{n=0}^\infty x^n=\sum_{n=0}^\infty \frac{d}{dx}x^n = \sum_{n=0}^\infty n x^{n-1}$$ and $$\frac{d^2}{dx^2}\frac{1}{1-x}=\sum_{n=0}^\infty\frac{d}{dx} n x^{n-1}=\sum_{n=0}^\infty n(n-1)x^{n-2}.$$ Hence $$-\frac{1}{2}\frac{d^2}{dx^2}\frac{1}{1-x} = -\sum_{n=0}^\infty \frac{n(n-1)x^{n-2}}{2} = 0+0-{1}{}-3x-\cdots.$$ Observe that the first two terms in the series expansion are zero and we can change the indexing of the series. In particular $$-\frac{1}{2}\frac{d^2}{dx^2}\frac{1}{1-x} = -\sum_{n=0}^\infty \frac{n(n-1)x^{n-2}}{2} = -\sum_{n=2}^\infty \frac{n(n-1)x^{n-2}}{2}.$$ This is equivalent to taking the original series (summing from zero to infinity) and replacing $n$ with $k+2$ (and summing over $k$ from $0$ to $\infty$) giving you the answer you're looking for!