Exhibit Function f Using Weierstrass Product Thm

[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.

So the solution goes as
Let $z_n$ be the nth term in the sequence $1,2,2,3,3,3,\ldots$.

Note that:
$$
\underbrace{\sum_{n=1}^{\infty}\frac{1}{|z_n|^3}}_{\text{Why to the 3rd power?}} = \underbrace{\sum_{n=1}^{\infty}\frac{n}{n^3}}_{ \text {Why is this equality true?}}
$$

 

[Dustinsfl]

Why is $k_n$ 3 here?

The Weierstrass Product is $\prod\limits_{n = 1}^{\infty}\left(1-\frac{z}{z_n}\right)^{-1}e^{-P\left(z/z_n\right)}$.

So that leaves us with $\displaystyle\prod\limits_{n = 1}^{\infty}\left(1-\frac{z}{z_n}\right)^{-1}e^{-P\left[\frac{z}{z_n}+\left(\frac{z}{z_n}\right)^2\right]}$

I was told that the above product can be simplified. Like this?

By expanding the product, we have
$$
\left(1 - z\right)^{-1}e^{\left[-z + \frac{z^2}{2}\right]}\left(1 - \frac{z}{2}\right)^{-2}e^{\left[-\frac{z}{2} + \frac{z^2}{4}\right]}\left(1 - \frac{z}{3}\right)^{-3}e^{\left[-\frac{z}{3} + \frac{z^2}{6}\right]}\cdots\left(1 - \frac{z}{n}\right)^{-n}e^{\left[-\frac{z}{n} + \frac{z^2}{2^n}\right]}\cdots
$$
So we have that
$$
\prod_{n = 1}^{\infty}\left(1 - \frac{z}{z_n}\right)^{-1}e^{-\left[\frac{z}{z_n} + \left(\frac{z}{z_n}\right)^2/2\right]} =
\prod_{n = 1}^{\infty}\left(1 - \frac{z}{n}\right)^{-n}\exp\left[{-\frac{z}{n} + \frac{z^2}{2n}}\right].
$$

 

[math771]

The reason for taking the 3rd power is because the Weierstrass Product Theorem involves the product of terms raised to a certain power, and in this case, we want the function $f$ to have a pole of order $n$ at each positive integer $n$. This means that the terms in the product should be raised to the power of $n$.

As for the equality, it is true because each term in the first sum is equal to $\frac{1}{n^3}$, while each term in the second sum is equal to $\frac{n}{n^3}$. Simplifying both sums, we get $\sum_{n=1}^{\infty}\frac{1}{n^2}$, which is a well-known convergent series. Therefore, the two sums are equal.