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:

$G \paren z = \dfrac 1 {\paren {1 - z}^2}$

Corollary

Let $\sequence {a_n}$ be the sequence defined as:

$\forall n \in \N_{> 0}: a_n = n - 1$

That is:

$\sequence {a_n} = 1, 2, 3, 4, \ldots$

Then the generating function for $\sequence {a_n}$ is given as:

$H \paren z = \dfrac z {\paren {1 - z}^2}$

Proof

Take the sequence:

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

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

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

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.

$\blacksquare$