Expansion of this expression Let $x$ be a real number in $\left[0,\frac{1}{2}\right].$ It is well known that 
$$\frac{1}{1-x}=\sum_{n=0}^{+\infty} x^n.$$
What is the expansion or the series of the expression $(\frac{1}{1-x})^2$?
Many thanks.

Or do by brute force multiplication and gathering terms:
$$\frac{1}{(1-x)^2}=$$
$$
(1+x+x^2+x^3+x^4+\ldots)(1+x+x^2+x^3+x^4+\ldots)\\
=(1+x+x^2+x^3+x^4+\ldots)\\
+x+x^2+x^3+x^4+\ldots\\
\quad +x^2+x^3+x^4+\ldots\\
\quad \quad +x^3+x^4+\ldots\\
=1+2x+3x^2+4x^3+\ldots
$$
