Generating Function for Natural Numbers

Theorem
Let $\left \langle {a_n}\right \rangle$ be the sequence defined as:
 * $\forall n \in \N_{> 0}: a_n = n$

That is:
 * $\left \langle {a_n}\right \rangle = 1, 2, 3, 4, \ldots$

Then the generating function for $\left \langle {a_n}\right \rangle$ is given as:
 * $\displaystyle G \left({z}\right) = \frac 1 {\left({1 - z}\right)^2}$

for $\left|{z}\right| < 1$.

Proof
Take the constant sequence:


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

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


 * $G_S \left({z}\right) = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$

for $\left|{z}\right| < 1$.

That is:
 * $G_S \left({z}\right) = 1 + x + x^2 + x^3 + \cdots$

Differentiating $G_S$ term by term:


 * $\displaystyle \frac {\mathrm d} {\mathrm d z} G_S = 0 + 1 + 2 x + 3 x^2 \cdots = \sum_{n = 0} \left({n + 1}\right) z^n$

which is the power series whose coefficients are $\left \langle {a_n}\right \rangle$.

But $G_S \left({z}\right) = \frac 1 {1 - z}$

and so by Power Rule for Derivatives and the Chain Rule:
 * $\displaystyle \frac {\mathrm d} {\mathrm d z} G_S = \frac 1 {\left({1 - z}\right)^2}$

The result follows from the definition of a generating function.