# Show convergence of infinite product

Let $$w \in \mathbb{C}$$ be a complex number with $$0 <|w|<1$$. Show that the infinite product $$\theta(z) := \prod_{n=1}^\infty (1+w^{2n-1}e^z)(1+w^{2n-1}e^{-z})$$ converges locally uniformly and absolutely for $$z \in \mathbb{C}$$.

We have a theorem that states that for a sequence of holomorphic functions $$f_n$$, if the infinite series of $$[f_n(z)-1]$$ converges absolutely and locally uniformly, then the infinite product of $$f_n(z)$$ converges absolutely and locally uniformly. I believe we can use this and solve the question by proving the convergence of the corresponding infinite series using the Weierstrass M-test, but I cannot seem to bound the series from above. Or how should I do this?

• Could you use the Jacobi triple product which relates it to an infinite sum? – Somos Apr 9 '20 at 18:31

## 1 Answer

Your series is $$\sum_{n=1}^\infty\left(\big|w^{2n-1}e^z\big|+\big|w^{2n-1}e^{-z}\big|\right)$$ For $$|w|<1$$ a fixed number and $$z$$ in a bounded set, this should be easy to bound.

Suppose $$z$$ is in the set $$\{z : |z| \le M\}$$. Then $$|e^z| = e^{\operatorname{Re} z} \le e^M$$ and $$|e^{-z}| = e^{-\operatorname{Re} z} \le e^M$$, so

$$\sum_{n=1}^\infty\left(\big|w^{2n-1}e^z\big|+\big|w^{2n-1}e^{-z}\big|\right) \le \sum_{n=1}^\infty\left(|w|^{2n-1}e^M+|w|^{2n-1}e^{M}\right) = \frac{2e^M |w|}{1-|w|^2} .$$ Therefore, by the Weierstrass M-test, the infinite series converges uniformly on the set $$\{z : |z|\le M\}$$. And so $$\theta(z)$$ converges uniformly and absolutely on $$\{z : |z| \le M\}$$.

• Unless I am mistaken, a term $w^{4n-2}$ is missing. – Martin R Apr 9 '20 at 16:23
• Corrected. Two terms in the sum correspond to two factors in the product. – GEdgar Apr 9 '20 at 21:26
• I can't seem to see how it bounds? Let's say we look in $B(0,1)$, then $|w^{2n-1}e^{±z}| \leq |e^{±z}| \leq e$, won't each term be bounded by the constant $e$, so that the series doesn't converge? – JjL7 Apr 9 '20 at 22:43
• Why isn't the last equality given by $\frac{2e^M}{ |w|} \cdot \frac{1}{1-|w|^2}$? – JjL7 Apr 10 '20 at 2:34
• The sum is from $n=1$ to $\infty$, not $n=0$ to $\infty$. – GEdgar Apr 10 '20 at 12:58