Riemann zeta function and uniform convergence

• Apr 9th 2013, 02:52 PM
gralla55
Riemann zeta function and uniform convergence
I've attached a picture of the problem.

This is a "challenge question" from my complex analysis book. I've seen the Riemann zeta function mentioned here and there, so I thought it could be fun to try to do something with it.

I tried looking it up on wikipedia, but there they just give this infinite sum as the "definition" of the function, which leads me to believe that this does in fact converge uniformly. However, using the limit definition of uniform convergence seems like a nightmare to do.

I'm therefore wondering if there is some obvious shortcut here or something. Any help would be appreciated!
• Apr 9th 2013, 06:52 PM
chiro
Re: Riemann zeta function and uniform convergence
Hey gralla55.

I'm just wondering if you can first compile a list of all equivalent statements of uniform convergence and then use the one that is most applicable?

This is a common technique when you have one thing that isn't as manageable, but through a duality or some other equivalence, it becomes more manageable.

I don''t know a lot about analysis myself so I can't give you a list of equivalent statements.
• Apr 9th 2013, 08:12 PM
hollywood
Re: Riemann zeta function and uniform convergence
You know what happens to the series at z=1, right?

- Hollywood
• Apr 9th 2013, 11:00 PM
gralla55
Re: Riemann zeta function and uniform convergence
Well, at z = 1, this just becomes the harmonic series? Which diverges but should it tell me something about uniform convergence? Thank you for your reply!
• Apr 10th 2013, 06:08 AM
xxp9
Re: Riemann zeta function and uniform convergence
for any given natural number N and any n>N, we can always find a real number s>1, such that |fn(s)-f(s)| > 1. Because you can choose the real number to be close to 1 enough.
• Apr 10th 2013, 06:31 AM
hollywood
Re: Riemann zeta function and uniform convergence
If the series diverges on the boundary of your region, it's not likely to converge uniformly inside the region. So how would you prove that it's *not* uniformly convergent?

I think xxp9 has the right idea.

- Hollywood
• Apr 10th 2013, 09:10 AM
gralla55
Re: Riemann zeta function and uniform convergence
Thank you both for your replies! Well, when I look at it that way, I'm now rather inclined to believe that the series does in fact converge uniformly. The series 1/n^2 converges uniformly, right? If so, shouldn't every series between 1/n and 1/n^2 also converge uniformly?
• Apr 10th 2013, 05:05 PM
hollywood
Re: Riemann zeta function and uniform convergence
To prove uniform convergence, you need an N such that $|f(x)-f_n(x)|<\epsilon$ for all $n>N$ and all x. The "and all x" is what makes the convergence uniform.

To disprove uniform convergence, you need to show that the definition can't be satisfied, so for a given $\epsilon$ and N, you need to find an x such that $|f(x)-f_n(x)|\ge\epsilon$.

There's no way to apply the definition of uniform convergence to $\sum_{n=1}^\infty\frac{1}{n^2}$ - you need a series of functions, not a series of numbers.

- Hollywood