I am recently self-studying Complex Analysis and came up with a question with regard to infinite products.
I am trying to show that $\prod_{k=1} {\cos(z/2^k)}$ converges. My first thought is to use the complex representation of cosine, $\sum (-1)^n z^2n/(2n)!$ But I just don't know how to get to the form $\prod (1+b_n)$ for which has a theorem for convergence.
Thanks for all the advice.
Hint. One may use a Taylor series expansion, observing that as $k \to \infty$, $$ \cos \frac{z}{2^k}=1-\frac{z^2}{2^{2k+1}}+O\left(\frac{z^4}{2^{4k}}\right). $$
Use$$\begin{align}\prod_{k=1}^\infty\cos\frac{z}{2^k}&=\prod_{k=1}^\infty\frac{\operatorname{sinc}\frac{z}{2^{k-1}}}{\operatorname{sinc}\frac{z}{2^k}}\\&=\lim_{N\to\infty}\prod_{k=1}^N\frac{\operatorname{sinc}\frac{z}{2^{k-1}}}{\operatorname{sinc}\frac{z}{2^k}}\\&=\lim_{N\to\infty}\frac{\operatorname{sinc}z}{\operatorname{sinc}\frac{z}{2^N}}\\&=\frac{\operatorname{sinc}z}{\lim_{N\to\infty}\operatorname{sinc}\frac{z}{2^N}}\\&=\frac{\operatorname{sinc}z}{\operatorname{sinc}0}\\&=\operatorname{sinc}z.\end{align}$$
