TITLE: convergence of $\prod_{n=1}^\infty (1-\frac{z}{n!})$
QUESTION [5 upvotes]: I want to show that $\prod_{n=1}^\infty (1-\frac{z}{n!})$ is convergent (or uniformly convergent) (z is complex)
Can I use the Theorem: 
The infinite product $\prod_{n=1}^{\infty} (1+a_n)$ converges if and only if the series $\sum_{n=1}^{\infty} a_n$ converges.

REPLY [3 votes]: Note that
\begin{align}
\lim_{N\to\infty}\left|\prod_{n=1}^{N}\left( 1-\frac{z}{n!}\right)\right|
=
&
\lim_{N\to\infty}\exp\left(\log\left(\prod_{n=1}^{N}\left| 1-\frac{z}{n!}\right|\right)\right)
\\
=
&
\lim_{N\to\infty}\exp\left(\sum_{n=1}^{N}\log\left| 1-\frac{z}{n!}\right|\right)
\\
\leq
&
\lim_{N\to\infty}\exp\left(\sum_{n=1}^{N}\log\left[1+\frac{|z|}{n!}\right]\right)
\\
\end{align}
Due to inequality $\log (x)<x-1$ for all $x>0$,
\begin{align}
\lim_{N\to\infty}\left|\prod_{n=1}^{N}\left( 1-\frac{z}{n!}\right)\right|
\leq
&
\lim_{N\to\infty}\exp\left(\sum_{n=1}^{N}\left[\frac{|z|}{n!}\right]\right)
\\
=
&
\lim_{N\to\infty}\exp\left(|z|\sum_{n=1}^{N}\left[\frac{1}{n!}\right]\right)
\\=
&
\exp\left(|z|\lim_{N\to\infty}\sum_{n=1}^{N}\left[\frac{1}{n!}\right]\right)
\\
=
&
\exp\left(|z|e\right)
\\
\end{align}