Derivative of Geometric Progression

Theorem

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

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

Corollary

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

Proof

We have from Power Rule for Derivatives that:

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

But from Sum of Infinite Geometric Progression:

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

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

$\blacksquare$

Proof of Corollary

We have from Power Rule for Derivatives that:

$\displaystyle \frac d {dx} \sum_{n \ge 1} \left({n+1}\right) x^n = \sum_{n \ge 1} n \left({n+1}\right) x^{n-1}$

But from Sum of Infinite Geometric Progression:

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

	\(\displaystyle \)	\(\displaystyle \sum_{n \ge 1} \left({n+1}\right) x^n\)	\(=\)	\(\displaystyle \sum_{m \ge 2} m x^{m-1}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{m \ge 1} m x^{m-1} - 1\)	\(\displaystyle \)	taking into account the first term
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {\left({1-x}\right)^2} - 1\)	\(\displaystyle \)	from main result above

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

$\blacksquare$

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): $\S 2.4$: Footnote