Exhibit Function f w/ Weierstrass Product Theorem

[Dustinsfl]

Use the Weierstrass Product Theorem to exhibit a function $f$ such that each positive integer $n$, $f$ has a pole of order $n$, and $f$ is analytic and nonzero at every other complex number

For $f$ to have a pole of order $n$, we have that $f = \prod\limits_{n = 1}^{\infty}\left(1 - \frac{z}{z_n}\right)^{-1}e^{-P_n(z/z_n)}$.
Let $z_n$ be the $\text{n}^{\text{th}}$ term in the sequence, i.e. $1, 2, 2, 3,\ldots$.
So taking $k_n$ to be 3, we have that (why is it 3?)
$$
\sum_{n = 1}^{\infty}\frac{1}{\left|z_n\right|^3} = \sum_{n = 1}^{\infty}\frac{1}{n^2} < \infty
$$
which converges since we have a p-series of degree two.
Now $P_n\left(\dfrac{z}{z_n}\right) = \dfrac{z}{z_n} + \dfrac{\left(\frac{z}{z_n}\right)^2}{2} + \cdots + \dfrac{\left(\frac{z}{z_n}\right)^{k - 1}}{k - 1}$ so the Weierstrass Product for $k_n = 3$ is

$$
\prod_{n = 1}^{\infty}\left(1 - \frac{z}{z_n}\right)^{-1}e^{-\left[\frac{z}{z_n} + \left(\frac{z}{z_n}\right)^2/2\right]}
$$

I was told that the above product can be simplified down. How?

 

[Dustinsfl]

So looking at the product, we have
$$
\left[\left(1-\frac{z}{1}\right)^{-1}\left(1-\frac{z}{2}\right)^{-2}\cdots\left(1-\frac{z}{n}\right)^{-n}\cdots\right]\exp\left[-z-\frac{z}{2}-\cdots -\frac{z}{n} - \cdots + \frac{z^2}{2^2}+ \frac{z^2}{6} + \frac{z^2}{8}\cdots\right]
$$

So can this be written as a different infinite product then from the final one I obtained?

All I see is
$$
\prod_{n=1}^{\infty}\left(1-\frac{z}{n}\right)^{-n}\exp\left[-\frac{z}{n}+\frac{z^2}{2^n}\right]
$$
but is this even the right observation?