Product with exponential converges absolutely and uniformly Prove that the product $$\prod_{n=1}^\infty \left(1+\frac{z}{n}\right)e^{-z/n}$$ converges absolutely and uniformly on every compact set.
What can I transform this product into? It's $\lim_{k\rightarrow\infty}\prod_{n=1}^k \left(1+\frac{z}{n}\right)e^{-z/n}$, but what to do with its convergence?
 A: By definition, the product converges (absolutely, uniformly) if and only if its logarithm
$$
\log \prod_{n=1}^\infty \bigg( 1+\frac zn\bigg) e^{-z/n} = \sum_{n=1}^\infty \bigg( \log\bigg(1+\frac zn\bigg) - \frac zn \bigg)
$$
converges (absolutely, uniformly).
The function $\log(1+w)-w$ has a double zero at $w=0$, and so $f(w) = (\log(1+w)-w)/w^2$ is analytic (setting $f(0)=-\frac12$) in a disk of radius $\frac12$ around $z=0$, say. In particular, it is continuous and hence bounded, so choose $M$ such that $|f(w)|\le M$ for $|w|\le \frac12$.
Now let $S$ be a compact subset of the complex numbers; there exists $B$ such that $|z|\le B$ for every $z\in S$. Let $T>2B$ be an integer. Then $\big|\frac zn\big| \le \frac12$ for every $z\in S$ and $n>T$, and so
$$
\sum_{n>T} \bigg| \bigg( \frac zn\bigg)^2 f\bigg( \frac zn \bigg) \bigg| \le \sum_{n>T} \frac{|z|^2}{n^2} M \le B^2M \sum_{n>T} \frac1{n^2} < B^2M \frac1T.
$$
Therefore the tail of our absolute-valued sum can be made small (less than $\epsilon$, say) by choosing $T>B^2M/\epsilon$; and this choice is independent of $z\in S$. Therefore the original sum converges absolutely and uniformly on $S$, hence so does the product.
