[SOLVED] show convergence of weierstrass product

[dwsmith]

$f(z) = \prod\limits_{n=1}^{\infty}\left(1+z^{2^n}\right)$ converges on the open disc $D(0,1)$ to the function $\dfrac{1}{1-z}$.

To show convergence, I look at
$$
\sum_{n=1}^{\infty}\left|z^{2^n}\right|
$$
correct?


The sum, $\sum\limits_{n = 0}^{\infty}|z|^{2^{n}}$, converges for $|z| < 1$ i.e. we have a geometric series.
So
$$
\sum_{n = 0}^{\infty}|z|^{2^{n}} = \frac{1}{1 - z},
$$
for $|z| < 1$

Is this convergence uniform on compact subsets of $D(0,1)$?

 

[dwsmith]

This should actually be done by the comparison test.

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

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?

 

[dwsmith]

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?).