Derivative of Geometric Sequence

Theorem
Let $x \in \R: \size x < 1$.

Then:


 * $\ds \sum_{n \mathop \ge 1} n x^{n - 1} = \frac 1 {\paren {1 - x}^2}$

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

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

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