# Order of an entire function represented as an infinite product

Question- What is the order of growth of the entire function given by the infinite product of $1-(z/n!)$ where $n$ goes from 1 to infinity?

My thoughts- I have already proven that the infinite product converges to an entire function. One can see that the zeros of the infinite product are at $n!$. So, by Hadamard's factorization theorem, I can deduce that the order $p$ is such that $0\leq p <1$.

I suspect the order of growth is zero, but using inequalities like $\lvert \log(1+z)\vert < \lvert z\rvert$ for $\lvert z\rvert < 1$ and $e^n < n!$ for $n\geq 6$, all I could get was the order of growth is less than $1$. I feel like there should be some simple inequality argument that shows us the order is zero.

Can somebody give me a hint? Thanks.

Claim: $p = 0$

Let $f(z) = \prod_{i=1}^\infty (1-\frac{z}{n!})$

I will show that for any $\epsilon >0$, the order of growth $p \leq \epsilon \;$ i.e. I will show $\exists \; A, B >0$ such that $|f(z)| \leq Ae^{B|z|^\epsilon}$.

\begin{align}|f(z)| = |\prod_{i=1}^\infty (1-\frac{z}{i!})| &= |\prod_{i! \leq |z|} (1-\frac{z}{i!})| \;|\prod_{i! > |z|} (1-\frac{z}{i!})| \\ &\leq \; \prod_{i! \leq |z|} (1+\frac{|z|}{i!}) \; \prod_{i! > |z|} (1+\frac{|z|}{i!}) \end{align}

Let us fix $z$, and suppose $\;n! \leq |z| < (n+1)!$ and $n\geq 6$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ (1)

Then, we use the inequality $e^n < n!$ for $n>6$ to get $e^n \leq |z| \implies n \leq log|z|$ $\;\;\;\;\;\;\;\;\;$ (2)

Now,

\begin{align}\prod_{i! \leq |z|} (1+\frac{|z|}{i!}) &\leq \prod_{i! \leq |z|} (\frac{2|z|}{i!}) \;\;\;(since \;\;\frac{|z|}{i!}\geq 1) \\ & \leq\prod_{i! \leq |z|} |z| \\ & \leq |z|^n \;\;\;\;\;\;\;\;\;\;\;(since \;\; n! \leq |z|) \\ & \leq |z|^{log|z|} \;\;\;\;\;\;\;\; by \;\mathbf{(2)} \\ \end{align}

Note that for any $\epsilon > 0, \frac{(log|z|)^2}{|z|^\epsilon} \rightarrow 0$ as $|z| \rightarrow \infty$.

Thus, $\exists \; C > 0, D > 0$ such that $$(log|z|)^2 \leq C + D|z|^{\epsilon} \implies |z|^{log|z|} \leq e^{C +D|z|^{\epsilon}}$$

Also,

\begin{align}\prod_{i! > |z|} (1 + \frac{|z|}{i!}) &= e^{\sum_{i! > |z|}(log(1+\frac{|z|}{i!}))} \\ &\leq e^{\sum_{i! > |z|} (\frac{|z|}{i!})} \\ &= e^{\frac{|z|}{(n+1)!}\sum_{i! > |z|} (\frac{(n+1)!}{i!})} \\ & \leq e^{\frac{2|z|}{n!}} \;\;\; (using \mathbf{(1)}) \\ & \leq e^2 \end{align}

If $|z| < 6!$, then $|f(z)|$ is bounded.

Thus, for any $\epsilon >0$, $\;$$\exists \; A, B >0$ $\;$ such that $|f(z)| \leq Ae^{B|z|^\epsilon}$.

Hence, $p = 0$

QED