# Uniform Convergence of a Complex Series

• January 1st 2011, 10:50 PM
roninpro
Uniform Convergence of a Complex Series
Hello. I'd like to show that the series

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{z-n}$

converges uniformly on any domain not containing an integer.

I'm not exactly sure how to handle this. It looks similar to an alternating harmonic series, so I can't use the $M$-test on it.

I would appreciate any suggestions for this problem.
• January 2nd 2011, 03:59 PM
Bruno J.
Quote:

Originally Posted by roninpro
Hello. I'd like to show that the series

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{z-n}$

converges uniformly on any domain not containing an integer.

I'm not exactly sure how to handle this. It looks similar to an alternating harmonic series, so I can't use the $M$-test on it.

I would appreciate any suggestions for this problem.

It's not true. For example, $\mathbb{C}-\mathbb{Z}$ is a domain not containing an integer, but the series doesn't converge uniformly there.

Perhaps you mean on any compact set not containing an integer?
• January 2nd 2011, 06:35 PM
roninpro

Suppose we do take the domains to be compact. I'm still not quite sure how to handle showing uniform convergence. Should I try to prove it directly from the definition?
• January 2nd 2011, 08:09 PM
Bruno J.
Well, a domain is a connected open set. A compact set in $\mathbb{C}$ must be closed (and bounded). The only compact domain is, in fact, the empty set! So you should probably forget completely about the "domain" part, and just replace "domain" by "compact set".

You should prove it directly from the definition. You might want to use the following theorem : if $C$ is a compact set, $E$ is closed, and $C\cap E = \emptyset$, then there exists $c\in C, e\in E$ such that $|c-e|$ is minimal and such that $|c-e|>0$. In other words, $C$ is a positive distance away from $E$.
• January 3rd 2011, 01:34 AM
Opalg
I think that the key additional condition you need is that the domain should be bounded (rather than compact).

Let $z = x+iy$. Then $\frac1{z-n} = \frac1{(x-n)+iy} = \frac{(x-n)-iy}{(x-n)^2+y^2}$, and so

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{z-n} = \sum_{n=1}^\infty \frac{(-1)^n(x-n)}{(x-n)^2+y^2} + i\sum_{n=1}^\infty \frac{(-1)^ny}{(x-n)^2+y^2}.$

Use the boundedness of the domain to show that each of those two sums converges. For the imaginary part, you can use the M-test (because the n'th term looks sort of like $1/n^2$). For the real part, show that $\frac{-(x-n)}{(x-n)^2+y^2}$ decreases monotonically and uniformly to 0 (for n suitably large), and use a uniform version of the alternating series test.