Infinite product convergence for cosine

Question

I have trouble proving the following: if $$\sum_k^\infty|z_k|^2 < \infty$$, then $$\prod_k^\infty \cos(z_k)$$ converges. (Note $z_k$ are complex numbers). I think some relevant proof of convergence is that if $\sum_{j=1}^\infty |1-a_j|<\infty$ then $\prod_{j=1}^\infty a_j$ converges. But not sure how to apply it here...

Answer 1

Strictly speaking, if $\cos z_k=0$ for some $k$, we have a zero product, and those are usually understood as divergent.

Suppose this does not happen. If $a_n\ne 0$ and $a_n\to 1$, the convergence of $\prod a_n$ is equivalent to convergence of $\sum |1-a_n|$, as you said. Another useful fact is the Taylor expansion of cosine at $0$, which yields $$ \lim_{z\to 0} \frac{1-\cos z}{z^2} =\frac12 $$ Hence, $\sum |z_k|^2$ and $\sum |1-\cos z_k|$ either both converge or both diverge, by the Limit Comparison test.