Infinite product $\prod \frac{R^{2^n}+z^{z^n}}{R^{2^n}}$ coverges

Question

Show the infinite product $$\prod_{n=1}^{\infty} \frac{R^{2^n}+z^{2^n}}{R^{2^n}}$$ converges for $|z|<R$, and $$\prod_{n=1}^{\infty} \frac{R^{2^n}+z^{z^n}}{R^{2^n}}=\frac{R}{R-z}$$ The converge is easy: check $$\sum log(\frac{R^{2^n}+z^{z^n}}{R^{2^n}})$$ But I don't know how to show the equality $$\prod_{n=1}^{\infty} \frac{R^{2^n}+z^{z^n}}{R^{2^n}}=\frac{R}{R-z}$$. Any suggestion is appreciated.

Answer 1

Assuming you meant $$\prod_{n=0}^{\infty}\frac{R^{2^n}+z^{2^n}}{R^{2^n}},$$ we have $$\prod_{n=0}^{\infty}\frac{R^{2^n}+z^{2^n}}{R^{2^n}}$$ $$=\prod_{n=0}^{\infty}\left(1+\left(\frac{z}{R}\right)^{2^n}\right)$$ $$= \sum_{S\subseteq \mathbb{N}}\left(\frac{z}{R}\right)^{\sum_{k \in S} 2^k}$$ $$= \sum_{n=0}^{\infty}\left(\frac{z}{R}\right)^n$$ $$=\frac{R}{R-z}.$$

The first equality comes from rewriting. The second comes from multiplying/distributing out the terms. The third comes from the fact that every natural number can be expressed as a sum of powers of 2 (i.e. in binary with 1s and 0s). The fourth comes from the geometric series.