# Derivative of Geometric Progression

From ProofWiki

## Contents

## Theorem

Let $x \in \R: \left|{x}\right| < 1$.

Then:

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

### Corollary

- $\displaystyle \sum_{n \mathop \ge 1} n \left({n+1}\right) x^{n-1} = \frac 2 {\left({1-x}\right)^3}$

## Proof

We have from Power Rule for Derivatives that:

- $\displaystyle \frac {\mathrm d} {\mathrm d x} \sum_{n \mathop \ge 1} x^n = \sum_{n \mathop \ge 1} n x^{n-1}$

But from Sum of Infinite Geometric Progression:

- $\displaystyle \sum_{n \mathop \ge 1} x^n = \frac 1 {1-x}$

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

$\blacksquare$

## Comment

Why should the more complicated second derivative result be a corollary of the (on the surface) easier first derivative result?

Because the *clever* bit is thinking of the differentiation technique in the first place. Applying it to the second result is just an exercise in application.

## Sources

- Geoffrey Grimmett and Dominic Welsh:
*Probability: An Introduction*(1986)... (previous)... (next): $\S 2.4$: Expectation: Example $24$: Footnote