Solve Gamelin's XIII.3 15: Infinite Product Meromorphicity

[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.

 

[Svein]

Take logarithms - it transforms the product into a sum.

 

[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 $$k>r$$.

 

[xiavatar]

I've posted a proof on http://math.stackexchange.com/quest...erges-to-meromorphic-function/1200018#1200018. I'm wondering if anyone can take a look at it and critique it.

 

[Svein]

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

 

[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?

 

[Svein]

we have to choose an analytic branch for the logarithm

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.

 

[xiavatar]

Okay, so we can choose an analytic branch, but your above equality with the logarithm

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

is only correct up to a multiple of $$2\pi i$$. What allows you to assume that the multiple vanishes in this case?

 

[Svein]

What allows you to assume that the multiple vanishes in this case?

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.