# Homework Help: Show converges uniformly on compact subsets of C

1. Feb 22, 2015

### Shackleford

1. The problem statement, all variables and given/known data

If α > 1, show: $∏ (1 - \frac{z}{n^α})$ converges uniformly on compact subsets of ℂ.

2. Relevant equations

We say that ∏ fn converges uniformly on A if

1. ∃n0 such that fn(z) ≠ 0, ∀n ≥ n0, ∀z ∈ A.

2. {∏ fn} n=n0 to n0+0, converges uniformly on A to a non-vanishing function g such that ∃δ > 0 with |g| ≥ δ.

Theorem: ∏ fn on A converges uniformly ⇔ ∀ε > 0, ∃n0 such that

$|∏ f_j(n) - 1| < ε$, ∀n ≥ m ≥ n0, and z ∈ A. converges uniformly on compact subsets of ℂ.

3. The attempt at a solution

Of course, the limit of z/nα is 0, and there exists an n0 such that for all larger n fn ≠ 0, i.e. when |z/nα| < 1.

Moreover, as n → ∞, each nth term in the term approaches 1, so the infinite product definitely converges, and I'd say that it satisfies the theorem given.

2. Feb 23, 2015

### RUber

If A is compact, does it have a least upper bound on |z|? That might be important to mention in your proof.

3. Feb 26, 2015

### Svein

This is correct with one caveat: The n0 you mention is dependent of z. Here you need the compactness: Since $C\subset \bigcup_{N} \left\{n>N\Rightarrow \frac{z}{n^{\alpha}}<\epsilon\right\}$, the compactness says that a finite number of these is sufficient. Take the maximum of these N, and you are done.

4. Mar 1, 2015

### Shackleford

Oh, yes. If it's compact, then it's close and bounded. Thanks for the refinement.