How to understand Weierstrass infinite product?

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One of the famous expression of $\sin(x)$ is

$\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{\pi^2n^2}\right)$

Here, we are writing $\sin(x)$ as a some product of polynomials with zeros apparently shown. Remember that the right hand side has to be carefully chose to ensure its convergence.

We are remained to consider: Can be factorize every entire function in the form of product of their zeros while maintaining its convergence? i.e. for any entire function, can be give the expression

$f(z)=g(z)f_0(z)f_1(z)\cdots$

where $f_0(a_0)=0$ , $f_1(a_1)=0\cdots$ , and $a_0, a_1\dots$ are the zeros.

If $f(z)$ is a polynomial, then the case is trivial. Here, we are only considering functions that cannot be expressed as polynomials.

Motivated by the expression of $\sin(x)$ , we could think of some possible constructions of factorizations.

We’ve expressed $\sin(x)$ as

$\sin(x)=\prod_n\left(1-\frac{z}{n\pi}\right)$

Notice that the RHS converges (proof will be left to the reader). However, does it converge generally for all

$\prod_n\left(1-\frac{z}{a_n}\right) ?$

We can prove that in some cases, this is not true.

To solve this issue, what we are supposed to do is to add additional factors that does not add any additional zeros. Apparently, the exponential function serves as a nice choice. What we want to do is to let $\exp(g(\frac{z}{a_n}))(1-\frac{z}{a_n})$ approaches to $1$ in an way faster than $(1-\frac{z}{a_n})$ . Actually, we could come up with this function

$E_k(z)=(1-z)e^{z+z^2/2+\cdots+z^k/k}$

In fact, this function $E_k$ is equivalent to

$e^{-\sum_{n=k+1}^{\infty}z^n/n}$

We could prove that this function is really close to $1$ on $|z|\leq \frac{1}{2}$ by showing that

$|1-E_k(z)|\leq c|z|^{k+1}$

Finally, we could show that for any function $f$ with given zeros $a_n$ with $|a_n|\rightarrow \infty$ , and it vanishes nowhere else. We have

$f(z)=z^m\prod_{n=1}^{\infty}E_k(z/a_n)$

By our previous lemma, we could divide zeros with $|a_n|\leq 2R$ and $|a_n|\geq 2R$ . For the first group, the product would satisfy the condition. When $|a_n|\geq 2R$ , we have $|z/a_n|<\frac{1}{2}$ when $z$ is inside the disk of radius $R$ . Thus, the product