Convergence of infinite product 2

Question

I have $\prod_{n=1}^\infty (1-e^{-\frac z n })$, and I must show that it converges in the points $z $ with real part greater than $0$. The sufficient condition doesn't seem to be satisfied, because $|-e^{-\frac z n }|=e^{-\frac z n }$, and since $e^{-\frac z n }\to 1$ as $n $ grows, the sum $\sum e^{-\frac z n }$ can't converge. Actually the factors approach $0$ as $n $ grows, namely they are asymptotic to $-\frac z n $. However this would suggest that the product converges to the entire function $f (z)=0 $, since $|\frac z n|\to 0 $ for all $z $. So probably this intuition is wrong (why?), since I had to prove that it converges in the right half of the complex plane. Thanks in advance for any correction.

Answer 1

$1-e^{-\frac z n} \to 0$ for every $z$. If $c_n \to 0$ then $\prod c_n=0$. To see this just note that $|c_n| <\frac 1 2$ for $n$ sufficiently large and $\prod \frac 1 2=0$. Hence the given product is $0$ for every complex number $z$ in the sense the partial products converge to $0$. [Some authors define such a product to be divergent].