I'm having difficulty showing the following identity (on open unit disk):
$$\frac{1}{1-z} = \prod_{n=0}^\infty{1+z^{2^n}}$$
My first idea was to show that they agreed on a set with non isolated points and show that the product converges to an analytic function on the unit disk, so that by uniqueness, they are equal. However, aside from 0, I cannot think of more appropriate values to choose. As well, I'm not quite sure how to show the uniform convergence of the infinite sum of |z|^(2^n) on compact subsets of the unit disk (or uniform convergence on unit disk?) to be able to show that the product converges to an analytic function.
Thank you!
Best Answer
Hint:
$$\frac{1}{1-z}=\prod_{n=0}^{\infty}(1+z^{2^n})$$
Recall that
$$\displaystyle \frac{1}{1-z}=\sum_{n=0}^{\infty}z^{n}$$
So, the equality
$$\prod_{n=0}^{\infty}(1+z^{2^n})=\frac{1}{1-z}=\sum_{n=0}^{\infty}z^{n}$$
boils down to the statement:
Try proving this by induction.
You will then have shown that these two objects are formally equal, and since they both converge on the disc, they are equal as functions.
EDIT: I was being silly--see Joriki's comment below.