Derivative of Geometric Sequence

Theorem
If $$x \in \R: \left|{x}\right| < 1$$, then:


 * $$\sum_{n \ge 1} n x^{n-1} = \frac 1 {\left({1-x}\right)^2}$$

Proof
We have from Power Rule for Derivatives that:
 * $$\frac d {dx} \sum_{n \ge 1} n x^{n-1} = \sum_{n \ge 1} x^n$$

But from Sum of Infinite Geometric Progression:
 * $$\sum_{n \ge 1} x^n = \frac 1 {1-x}$$

The result follows by Power Rule for Derivatives and the Chain Rule applied to $$\frac 1 {1-x}$$.