# convergence of $\prod_{n=1}^\infty (1-\frac{z}{n!})$

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.

• That's exactly what you need Commented Jan 13, 2015 at 23:05
• @user134824 But I think, $a_n$ is real in theorem but I have complex series.. I confused.. Commented Jan 13, 2015 at 23:09
• The theorem you are looking at may only be stated for real numbers, but it is true for complex numbers as well Commented Jan 13, 2015 at 23:11

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}