Derivative of Generating Function/General Result

Theorem
Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{a_n}\right\rangle$. Let $m$ be a positive integer.

Then:


 * $\dfrac {\d^m} {\d z^m} \map G z = \displaystyle \sum_{k \mathop \ge 0} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k$

Proof
Proof by induction:

Base Case
When $n = 0$, we have:
 * $\dfrac {\d^0} {\d z^0} \map G z = \map G z$

Also:
 * $\displaystyle \sum_{k \mathop \ge 0} \dfrac {\paren {k + 0}!} {k!} a_{k + 0} z^k = \sum_{k \mathop \ge 0} a_k z^k$

This is our base case.

Induction Hypothesis

 * $\forall n \in \N: n \ge 0: \dfrac {\d^n} {\d z^n} \map G z = \displaystyle \sum_{k \mathop \ge 0} \dfrac {\paren {k + n}!} {k!} a_{k + n} z^k$

This is our induction hypothesis.

It is to be demonstrated that:
 * $\dfrac {\d^{n + 1} } {\d z^{n + 1} } \map G z = \displaystyle \sum_{k \mathop \ge 0} \dfrac {\paren {k + n + 1}!} {k!} a_{k + n + 1} z^k$

Induction Step
This is our induction step:

Consider $m = n + 1$.

Hence the result by induction.