Derivative of Geometric Sequence/Corollary

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

Then:
 * $\ds \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1} = \frac 2 {\paren {1 - x}^3}$

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

But from Sum of Infinite Geometric Sequence:

The result follows by Power Rule for Derivatives and the Chain Rule for Derivatives applied to $\dfrac 1 {\paren {1 - x}^2}$.