Infinite product converges to meromorphic function How do you show that $\frac{1}{z}\prod_{n=1}^\infty \frac{n}{z+n}(\frac{n+1}{n})^z$ is meromorphic? Any hints would be helpful, I'm having trouble bounding the functions and their logarithms. This is exercise XIII.3 problem 15 in Gamelin's Complex Analysis. I have tried to prove it using the fact that $\prod_{n=1}^\infty (1+a_n)$ converges uniformly if $\sum_{n=1}^\infty |a_n|$ converges uniformly. I will post it below and see if anyone feels that it's okay.

So to show that our product is meromorphic its sufficient to show that it converges to a meromorphic function on closed disks of arbitrary radius $r$. Let $k>r$, so that $z+k$ does not vanish. We want to show if $\prod_{n=k}^\infty \frac{n}{z+n}(\frac{n+1}{n})^z$ converges to a holomorphic function on the closed disk $\overline{\mathbb{D}}_r$ centered at zero. It is thus sufficient to show that $\sum_{n=k}^\infty |\frac{n}{z+n}(\frac{n+1}{n})^z-1|$ converges uniformly. But the below series of estimations gives us 
$$\begin{align}
\sum_{n=k}^\infty |\frac{n}{z+n}(\frac{n+1}{n})^z-1|&\leq\sum_{n=k}^\infty |\frac{n}{z+n}||(\frac{n+1}{n})^z-1-\frac{z}{n}|\\
&\leq \sum_{n=k}^\infty |\frac{n}{z+n}||(\sum_{m=0}^\infty\frac{z\text{ log}(1+\frac{1}{n})^m}{m!})-1-\frac{z}{n}|\\
&\leq\sum_{n=k}^\infty |\frac{n}{z+n}||(\sum_{m=0}^\infty\frac{z^m}{m!n^m})-1-\frac{z}{n}|\\
&\leq\sum_{n=k}^\infty |\frac{n}{z+n}||(\sum_{m=2}^\infty\frac{z^m}{m!n^m})-1-\frac{z}{n}|\\
&\leq\sum_{n=k}^\infty |\frac{n}{z+n}||\frac{z^2}{n^2}||(\sum_{m=2}^\infty\frac{z^m}{m!n^m})|\\
&\leq K\sum_{n=k}^\infty\frac{1}{n^2}
\end{align}$$
where $K$ is a constant.Thus, the product $\prod_{n=k}^\infty \frac{n}{z+n}(\frac{n+1}{n})^z$ converges uniformly on $\mathbb{D}_r$. Hence is holomorphic. And therefore, the product $\frac{1}{z}\prod_{n=1}^\infty \frac{n}{z+n}(\frac{n+1}{n})^z$ converges to a meromorphic function on $\overline{\mathbb{D}}_r$. Convergence on compact sets then implies that we have a meromorphic function on all of $\mathbb{C}$.
