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.

Given that $$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$ for all $x\in(-1,1)$, we may differentiate both sides to obtain
$$\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty}nx^{n-1} = \sum_{n=0}^{\infty}(n+1)x^n$$
for all $x\in(-1,1)$.
Here we have used the fact that a power series is differentiable inside it's interval of convergence, and the derivative can be obtained by differentiating term-by-term, just as you would for a polynomial.
