About the proof of Zeta Transform

Question

I have to prove $$Z[k^n]=(-1)^nD^n\left(\frac{z}{z-1}\right)$$ where $$D=z\frac{d}{dz}$$ and $n$ varies over the set $\mathbb{Z}$ My book doesn't give me any advice; how can I go further?

Answer 1

Define:

$$f_n(z)=Z[k^n]=\sum_{k=0}^\infty k^n z^{-k}$$

Prove it by induction on $n$.

If $n=0$, then $$f_0(z)=\sum_{k=0}^\infty z^{-k} = \frac{1}{1-z^{-1}}=\frac{z}{z-1} = (-1)^0D^0\left(\frac{z}{z-1}\right)$$

Next, show that:

$$f_{n+1}(z)=(-1)D[f_n(z)]$$

by:

$$\begin{align}(-1)D[f_n(z)]&=(-1)D\left[\sum_{k=0}^\infty k^n z^{-k}\right] \\&= (-1)z\sum_{k=0}^\infty (-1)k^{n+1} z^{-k-1}\\&=f_{n+1}(z) \end{align}$$

More generally:

$$Z[ka_k] = (-1)D\left(Z[a_k]\right)$$

This is the property called differentiation on the wikipedia page for the Z-transform.