For $\frac{\sin(\pi z)}{\pi z} =\prod\limits_{n =1}^\infty \left(1-\frac{z^2}{n^2}\right)$, prove convergence of $\prod\limits_{n =1}^\infty \left(1-\frac{z^2}{n^2}\right)$ for any $z \in \mathbb{C}$.
$$f(x):=\ln\left(\frac{\sin(\pi z)}{\pi z}\right) =\sum\limits_{n =1}^\infty \ln\left(1-\frac{z^2}{n^2}\right) $$
Applying the root test for the $n$-th term we get the limit $R:=\lim\limits_{n \to \infty} \sqrt[n]{\ln\left(1-\frac{z^2}{n^2}\right)} $
$$\ln(R)=\lim\limits_{n \to \infty} \frac{\ln\left(\ln\left(1-\frac{z^2}{n^2}\right)\right)}{n} $$
By L'Hôpital's rule:$$\ln(R)=\lim\limits_{n \to \infty} \frac{2}{n}\frac{\frac{z^2}{n^2}}{\ln\left(1-\frac{z^2}{n^2 }\right)\left(1-\frac{z^2}{n^2}\right)}$$
As $n \to \infty $, $\left(1-\frac{z^2}{n^2}\right) \to 1$ and since $\lim\limits_{x \to 0}\frac{\ln(1-x)}{x}=-1$
So $\ln(R)=0$ so $R=1$.
But this produnct should converge for every $z$ (Wolfram gives values to all $z:|z|>1$ I tried) So unless I made a mistake (I usually make a lot of small errors and spent hours to find them ) the product only make since when $|z|<1$, If that is true then it could converge for some $|z|=1$ but this is not the problem!