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**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**

- Point on unit circle such that $|(z-a_1)\cdots(z-a_n)|=1$
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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?

- Is $z^{-1}(e^z-1)$ surjective?
- Zero divided by zero must be equal to zero
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- Proving the set $\mathcal H$ of Möbius transformations is a group under composition and finding a transformation that satisfies certain conditions
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- What function can be differentiated twice, but not 3 times?
- Is this proof of the fundamental theorem of calculus correct?

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.

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