Weierstrass Product convergence

[Dustinsfl]

Show that the infinite product f(z) = \prod\limits_{n = 0}^{\infty}(1 + z^{2^n}) converges on the open disc D(0,1) to the function 1/(1 - z). Is this convergence uniform on compact subsets of the disc?



This should actually be done by the comparison test.

For |z| < 1, we have that
$$
\frac{1}{1-z}=\sum_{n=0}^{\infty}|z|^{2^n}\leq \sum_{n=0}^{\infty}|z|^n
$$

So now I need to show by partial products that there are no zeros in K\subset D(0,1). So I need to find an N such that forall n > N this holds. I am looking for some guidance on this piece.

$$
f(z) = \prod_{n=0}^{N}\left(1+z^{2^n}\right) \prod_{n=N+1}^{\infty}\left(1+z^{2^n}\right)
$$

We need to fix R\in\mathbb{R}^+. Let N\in\mathbb{N} such that |z_N|\leq 2R < |z_{N+1} (is this correct-the inequalities?).

The first partial product is finite on D(0,1) and the second partial product behaves well on D(0,1).

What would be my choice of k_n for this product?

 

[Dustinsfl]

So continuing.

Then for |z|\leq R and n>N we have
$$
\left|\frac{z}{z_n}\right| <\frac{1}{2}, \quad\forall n>N
$$
so by Lemma: If |z|\leq 1/2, then \log\left[\prod\limits_{n=1}^{\infty}\left(1-\frac{z}{z_n}\right)\right]\leq 2|z|^n,
$$
\log\left|\left[\prod\limits_{n=1}^{\infty}\left(1-z^{2^n}\right)\right]\right| = \sum_{n=1}^{\infty}\left|\log(1+z^{2^n})\right| \leq 2\left(\frac{R}{z_n}\right)^{k_n}.
$$

What choices of $k_n$ will allow convergence(uniform/absolute?).