Limit of $(1+z)(1+z/2)\cdots(1+z/n)$ 
Let $z$ be a complex number such that $\operatorname{Re}(z)<0$ and
$$z_n = (1+z)\Bigl(1+\frac{z}{2}\Bigr) \cdots \Bigl( 1+\frac{z}{n} \Bigr).$$
Prove that $ \lim_{n \to \infty} z_n = 0$.

I noticed that: If $| z_n |^2 \to 0$ then $z_n \to 0$.
\begin{align*}
\lvert z_n \rvert^2
&= z_n \overline{z_n} \\
&= (1+z)(1+\overline{z}) \cdot \Bigl(1+ \frac{z}{n}\Bigr) \Bigl(1+ \frac{\overline{z}}{n}\Bigr) \\
&= (1 + 2 \operatorname{Re}(z) + \lvert z \rvert^2) \cdots \biggl(1 + \frac{2\operatorname{Re}(z)}{n} + \frac{\lvert z \rvert^2}{n^2}\biggr).
\end{align*}
For a fixed $z$ there will be such $ n \in \mathbb{N}$ such that for every $m >n$  we have $\frac{2\left|\operatorname{Re}(z)\right|}{m} > \frac{\left|z\right|^2}{m^2}$ .
Which means $$ 1 + \frac{2\operatorname{Re}(z)}{m} + \frac{\lvert z \rvert^2}{m^2} < 1.$$
So from one point we will be multiplying by a factor always smaller than $1$ (but closer and closer to $1$).
Is that a proper approach to the problem? I understand it is not enough to conclude that the limit is $0$.
 A: Write
$$ z_n = \exp\biggl( z H_n + \sum_{k=1}^{n} \left[ \log\left(1+\frac{z}{k}\right) - \frac{z}{k} \right] \biggr), $$
where $H_n = \sum_{k=1}^{n} \frac{1}{k}$ is the $n$th harmonic number. Since $ \left| \log(1+w) - w \right| \leq 4\left|w\right|^2 $ for all $\left|w\right|\leq\frac{1}{2}$, we find that
$$ \sum_{k=1}^{\infty} \left[ \log\left(1+\frac{z}{k}\right) - \frac{z}{k} \right] $$
converges. This shows that
$$ \left| z_n \right| \leq e^{\operatorname{Re}(z_n)H_n + C} $$
for some finite constant $C$ whose value is not so important to us. Since $H_n \to \infty$ as $n\to\infty$, it follows that $\left|z_n\right| \to 0$.
A: $$\log \left(\prod _{k=1}^n \left(\frac{z}{k}+1\right)\right)=\sum _{k=1}^n \log \left(\frac{z}{k}+1\right)$$
as $k\to \infty$ we have $\log \left(\frac{z}{k}+1\right)\to \frac{z}{k}$
and $$\sum _{k=1}^n \frac{z}{k}=z H_n$$
$H_n$ being the $n$th harmonic number.
$\text{Re}(z)<0$ therefore $\text{Re}(z H_n)\to-\infty$ and so $\text{Re}\left(\prod _{k=1}^n \left(\frac{z}{k}+1\right)\right)\to 0$ as $n\to\infty$
Hope this can be useful
