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$f(z) = \prod\limits_{n=1}^{\infty}\left(1+z^{2^n}\right)$ 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 $D(0,1)$?

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 $D(0,1)$?

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