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?