Prove that a power series $\sum_{n=0}^{\infty}{{a_n}{z^n}}$ which converges for any $z \in \mathbb{N}$, converges for any $z \in \mathbb{C}$.

PROBLEM

Prove that a power series $\sum_{n=0}^{\infty}{{a_n}{z^n}}$ which converges for any $z \in \mathbb{N}$, converges for any $z \in \mathbb{C}$.

MY ATTEMPT

Since the power series
$$\sum_{n=0}^{\infty}{{a_n}{z^n}}$$
converges for any $z \in \mathbb{N}$, then
$$\sum_{n=0}^{\infty}{{a_n}{z^n}}$$
can be considered (?) to be a polynomial in $\mathbb{C}$.

Hence,
$$\sum_{n=0}^{\infty}{{a_n}{z^n}}$$
is entire, and therefore continuous in $\mathbb{C}$ (since differentiability implies continuity).

It follows that
$$\sum_{n=0}^{\infty}{{a_n}{z^n}}$$
converges for any $z \in \mathbb{C}$.

QUESTION

I am aware that, while I may have (some) ideas for a (general) proof, admittedly there are several grey areas (i.e., gaps) that need to be filled. Would you be kind enough as to fill in these gaps, by providing useful hints?

Solutions Collecting From Web of "Prove that a power series $\sum_{n=0}^{\infty}{{a_n}{z^n}}$ which converges for any $z \in \mathbb{N}$, converges for any $z \in \mathbb{C}$."

If the radius of convergence of $\sum_{n=1}^{\infty}a_nz^n$ were finite, then there would exist $R\in [0,\infty)$ such that $\sum_{n=1}^{\infty}a_nz^n$ diverges for all $z$ with $|z|>R.$ Since this is not the case with our given series, the radius of convergence is $\infty,$ so it converges for all $z.$

This a consequence of a corollary Abel’s lemma:

If $f(z) = \sum a_n z^n$ is a power series and if the sequence $(|a_n|s^n)$ is bounded then $f$ converges absolutely in $B(0,s)$.

Proof Indeed,
if $0<t<s$ then $q=t/s <1$ and $|a_n|t^n \leqslant M q^n$ where $M$ is the bound we’re assuming exists, so
$$\sum |a_n|t^n \leqslant M \frac{1}{1-q}<\infty$$

The corollary: notice now that if a powerseries converges at a natural number $k$ (or just any complex number), then the sequence $(|a_n|k^n)$ tends to $0$ and is certainly bounded.