# Infinite product

1. Mar 21, 2015

### xiavatar

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.

2. Mar 21, 2015

### Svein

Take logarithms - it transforms the product into a sum.

3. Mar 21, 2015

### xiavatar

I did take logarithms, but the problem is bounding it and applying Weirstrass M-test. So what were reduced to showing is that $$\sum_{n=k}^\infty \text{log}(\frac{n}{z+n}(\frac{n+1}{n})^z)\leq \sum_{n=k}^\infty M_n<\infty$$ on a disk $$D_r$$ centered at zero where $$k>r$$.

Last edited: Mar 21, 2015
4. Mar 21, 2015

### xiavatar

5. Mar 22, 2015

### Svein

$log(\frac{n}{z+n}(\frac{n+1}{n})^{z})=log(n)-log(z+n)+zlog(n+1)-zlog(n)=(1-z)log(n)-log(z+n)+zlog(n+1)$.

6. Mar 22, 2015

### xiavatar

But the problem is that complex logarithms don't preserve the properties were used to with real valued logarithms. In order for them to be holomorphic we have to choose an analytic branch for the logarithm, and once we've done that we can't just rewrite the log as you have written. Or have I misunderstood something?

7. Mar 22, 2015

### Svein

Yes, but we have some help. I propose that we go for the branch where the argument is between -π and π. This is trivially fulfilled for log(n) and log(n+1). For sufficiently large n, the argument for log(z+n) is even between -π/2 and π/2.

Last edited: Mar 22, 2015
8. Mar 22, 2015

### xiavatar

Okay, so we can choose an analytic branch, but your above equality with the logarithm
is only correct up to a multiple of $$2\pi i$$. What allows you to assume that the multiple vanishes in this case?

9. Mar 23, 2015

### Svein

Well, since I am only doing logarithms in order to get a grip on the expression, I don't much care. The original expression is not in logarithms, so we need to take the exponent anyway.