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.

By the Cauchy product,
$$\left(\sum_{n=0}^{\infty} x^n\right)^2=\left(\sum_{n=0}^{\infty} x^n\right)\left(\sum_{m=0}^{\infty} x^m\right)=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}x^nx^m=\sum_{k=0}^{\infty}\sum_{m=0}^k x^k=\sum_{k=0}^{\infty}(k+1)x^k.$$
