Weierstrass Factorization Theorem, infinite polynomial/infinite power series

It's a bit complicated.

Power series

$$P(z) = \sum_{k=0}^{\infty} c_kz^k$$ and the more general Laurent series $$L(z) = \sum_{k=-\infty}^{\infty} c_kz^k$$ have a radius of convergence. Their values outside of this radius are not defined.

In analysis of one complex variable we learn that we can always construct a power series or Laurent series around a point $z_0$ so that its radius of convergence precisely touches the nearest pole.

A famous example we learn in calculus (one real variable) that the McLaurin series of $\arctan(t)$ converges on $[-1,1]$ on the real line. In one complex variable we can prove that this happens because

1. center point is $z=0$
2. $\arctan(z) = \frac{i}2 \left(\log(1-iz) - \log(1+iz)\right)$
3. closest poles to $z=0$ are $z=\pm i$
4. $\|\pm i\| = 1$

Now any partial sum of power series will have as many zeroes as it's degree and we can factor any such partial sum with Weierstrass Factorization theorem.

But partial sums to power series are not unique.

Because of the fundamental theorem of Algebra, somewhere outside of the radius of convergence there will be hopping around excess zeroes in some pattern for our choice of partial sum as we increase it's number of terms.

Let $E_0(z)=1-z, E_n(z)=E_0(z)\exp(z+z^2/2+..z^n/n)$ the Weierstrass factors and let $f$ an analytic function in some domain $G$

Then:

1: If $G=\mathbb C$ and $z_1,z_2...$ is the possibly empty set of zeroes of $f$ (repeated for mutliplicity) excluding $0$, there is an entire function $g$ and non-negative integers $k_0,k_1,..$ st:

$f(z)=e^{g(z)}z^{k_0}E_{k_1}(z/z_1)...E_{k_n}(z/z_n)...$ with normal convergence in the plane (the partial products converge absolutely on every compact subset of the plane)

The product can, of course, be empty or finite and is not generally unique; in various cases (eg finite order) one can have unicity but even simple examples like $\sin \pi z$ show that flexibility is better than unicity since one prefers the product with factors $1-z^2/n^2$ or if one wants the conditional product obtained by grouping $n$ with $-n$ to the Weierstrass factors of degree $1$ (whose exponential part cancels out if we group like that)

2: If $G \ne \mathbb C$ is simply connected and $z_1,z_2...$ is the possibly empty set of zeroes of $f$ (repeated for mutliplicity), there are $w_1,w_2,...$ outside $G$, $k_1,k_2,..$ non-negative integers and $g \in Hol(G)$ s.t.

$f(z)=e^{g(z)}E_{k_1}(\frac{z_1-w_1}{z-w_1})...E_{k_n}(\frac{z_n-w_n}{z-w_n})...$ with normal convergence in $G$ (the partial products converge absolutely on every compact subset of the given simply connected domain)

The product can, of course, be empty or finite and is not generally unique; in various cases (eg bounded or more generally Hardy space functions on a disc) one can have unicity (Blaschke products) and even more refined decomposition of the non-zero term (outer functions, singular inner functions etc)

3: If $G$ is now not simply connected the above holds but with some $h$ (instead of $e^{g(z)}$) in the group of invertible holomorphic functions on $G$ ($1/h$ holomorphic also or $h$ has no zeroes there); now $h$ is not anymore an exponential in general (see $1/z$ on the punctured plane or disc)

Since power series have a simply connected domain of definition (a possibly infinite radius disc), the result holds as long as they have a non-zero radius of convergence; if they satisfy reasonable growth conditions at the boundary (or at infinity if entire) then we have some form of unicity (Blaschke products and inner-outer decomposition or finite order Weierstrass products where $g$ is a polynomial respectively)

Edit later: As per comments, regarding proof this is a classic result and the proof takes a while, though the idea is simple - using the discreteness of the zeroes (so, in particular, the fact that they are at most countable and can be numbered for example in increasing absolute value in the entire case) and the property of the Weierstrass product $E_n(z)=1-\sum_{k \ge n+1}a_kz^k, a_k \ge 0, \sum a_k=1$, construct $F$ a function having the zeroes of $f$, then $f/F$ will have no zeroes so will be an exponential in the simply connected case; Remmert Classical Topics in Complex Function Theory has a very good proof in Chapters 3 and 4 with historical references and lots of insight; Rudin's RCA has a straightforward (but less enlightening) proof in Chapter 15 so will refer you to these and probably any good book on introductory complex analysis for that.