This is an exercise given to us in our analytic number theory class:
Prove that $ \prod \cos(\tfrac{z}{2^n}) $ is uniformly and absolutely convergent on every closed disk $ \{ |z| \leq R \} $, hence its value $ P(z) $ is an analytic function on $ \mathbb{C} $. Evaluate $ P(z) $ in simple terms.
We don't have a textbook for the course. I think I need to use the Weierstrass M-test here but I am not sure how to proceed. Any suggestions?
Let $P_n(z)=\prod_{k=1}^n \cos\left(\frac{z}{2^k}\right)$. Inductively we have \begin{align} P_1(z)\cdot \sin\left(\frac{z}{2}\right)&=\cos\left(\frac{z}{2}\right)\sin\left(\frac{z}{2}\right)=\frac{1}{2}\sin z,\\ P_2(z)\cdot\sin\left(\frac{z}{2^2}\right)&=\cos\left(\frac{z}{2}\right)\cos \left(\frac{z}{2^2}\right)\cdot\sin\left(\frac{z}{2^2}\right)\\ &=\frac{1}{2}\cos\left(\frac{z}{2}\right)\sin\left(\frac{z}{2}\right)=\frac{1}{2^2}\sin z ,\\ &...\\ &...\\ P_n(z)\cdot\sin\left(\frac{z}{2^n}\right)&=P_{n-1}(z)\cos\left(\frac{z}{2^n}\right)\sin\left(\frac{z}{2^n}\right)\\ &=\frac{1}{2}P_{n-1}(z)\sin\left(\frac{z}{2^{n-1}}\right)=...=\frac{1}{2^n}\sin z . \end{align} Therefore we have $$ P_n(z)=\frac{1}{2^n}\frac{\sin z}{\sin\left(\frac{z}{2^n}\right)}=\frac{\sin z}{z}\cdot\frac{\frac{z}{2^n}}{\sin\left(\frac{z}{2^n}\right)}\to \frac{\sin z}{z}\quad (\, n\to \infty\,) .$$ Thus, in simple terms, $P(z)=\frac{\sin z}{z}.$
Uniform and absolute convergence:
We can rewrite $$\prod_{n=1}^\infty \cos\left(\frac{z}{2^n}\right)=\prod_{n=1}^\infty (1+a_n),$$ where $a_n=\frac{1}{2}\left\{\left(e^{i\frac{z}{2^n}}-1\right)+ \left(e^{-i\frac{z}{2^n}}-1\right) \right\}.$
We have for $|z|\le R$ that \begin{align} \left|e^{\pm i\frac{z}{2^n}}-1\right|&=\left|\frac{1}{1!}\left( \pm i\frac{z}{2^n} \right)+\frac{1}{2!}\left( \pm i\frac{z}{2^n} \right)^2+\cdots \right|\\ &\le \left|\frac{1}{1!}\left( \frac{R}{2^n} \right)+\frac{1}{2!}\left( \frac{R}{2^n} \right)^2+\cdots \right|\\ &=e^{\frac{R}{2^n}}-1\\ &\le \frac{e^R-1}{2^n}, \end{align} since $e^x-1\le \frac{e^R-1}{R}\,x$ for $0\le x\le R.$
Therefore we have $|a_n|\le \frac{e^R-1}{2^n}$ which ensures uniform and absolute convergence of $\sum_{n=1}^\infty a_n$.