$\prod_{n=1}^{\infty}(1+\frac{z}{n^2})$ converges absolutely $\forall z \in \mathbb{C}$

Question

$$\prod_{n=1}^{\infty}(1+\frac{z}{n^2})$$ converges absolutely $\forall z \in \mathbb{C}$

I am told that $$\prod_{n=1}^{\infty}(1+z_n)$$ absolutely converges if $$\prod_{n=1}^{\infty}(1+|z_n|)$$ converges (Im not sure if I wrote this correctly though..) Anyway, the answer is given as: $$\prod_{n=1}^{\infty}(1+\frac{|z|}{n^2}) < \infty\iff \sum_{n=1}^{\infty}\frac{|z|}{n^2}< \infty $$ Why, and how is this equivalent??

Answer 1

The Weierstrass product for the sine function $$\frac{\sin z}{z}=\prod_{n\geq 1}\left(1-\frac{z^2}{\pi^2 n^2}\right) \tag{1}$$ is uniformly convergent over any compact subset of $\mathbb{C}$, hence it follows that $$ \prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right) = \frac{\sinh(\pi z)}{\pi z}\tag{2} $$ for any $z\in\mathbb{C}$.