How to solve this summatory?: $$\sum _{n=0}^{\infty }n^2\:0.999^{n-1}\cdot 0.001$$
I know how to calculate this one: $$\sum _{n=0}^{\infty }n\:0.999^{n-1}\cdot 0.001$$ as a sumatory of an arithmetic-geometric progression, but I don't know how to calculate this one with the square.
Call $$ f(x):=\sum_{n=0}^{+\infty}x^n=\frac{1}{1-x}\;\;,\;|x|<1. $$ Clearly $$ f'(x)=\sum_{n\ge1}nx^{n-1}=\frac{1}{(1-x)^2}\;\;. $$ Then \begin{align*} f''(x) &=\sum_{n=2}^{+\infty}n(n-1)x^{n-2}\\ &=\sum_{n=2}^{+\infty}n^2x^{n-2}-\sum_{n=2}^{+\infty}nx^{n-2}\\ &=\frac1x\left(\sum_{n=0}^{+\infty}n^2x^{n-1}-\sum_{n=0}^{+\infty}nx^{n-1}\right) \end{align*} from which you get \begin{align*} \sum_{n=0}^{+\infty}n^2x^{n-1} &=xf''(x)+\sum_{n=0}^{+\infty}nx^{n-1}\\ &=\frac{2x(1-x)}{(1-x)^4}+\frac1{(1-x)^2}\\ &=\frac{1-x^2}{(1-x)^4}. \end{align*}
