# 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.