Generating Function for Natural Numbers

Theorem
Let $\sequence {a_n}$ be the sequence defined as:
 * $\forall n \in \N_{> 0}: a_n = n - 1$

That is:
 * $\sequence {a_n} = 0, 1, 2, 3, 4, \ldots$

Then the generating function for $\sequence {a_n}$ is given as:
 * $G \paren z = \dfrac 1 {\paren {1 - z}^2}$

Proof
Take the sequence:


 * $S_n = 1, 1, 1, \ldots$

From Generating Function for Constant Sequence, this has the generating function:


 * $\ds G \paren z = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$

By Derivative of Generating Function:


 * $\ds \dfrac \d {\d z} G \paren z = 0 + 1 + 2 z + 3 z^2 \cdots = \sum_{n \mathop \ge 0} \paren {n + 1} z^n$

which is the generating function for the sequence $\sequence {a_n}$.

But:
 * $G \paren z = \dfrac 1 {1 - z}$

and so by Power Rule for Derivatives and the Chain Rule for Derivatives:
 * $\dfrac \d {\d z} G \paren z = \dfrac 1 {\paren {1 - z}^2}$

The result follows from the definition of a generating function.