Prove that exponential series squared is continous

Question

Prove that $$f(z)=\sum_{k=0}^{\infty} \left( \frac{z^k}{k!} \right)^2, z \in \mathbb{C} $$ is well-defined and continous.

I'm aware of the Weierstrass M-test, but I failed to find a proper estimation to apply it.

Thanks in advance.

Answer 1

It suffices to show that the interval of convergence is all of ${\bf{C}}$. By ratio test, we have \begin{align*} \lim_{k\rightarrow\infty}\left|\dfrac{\left(\dfrac{z^{k+1}}{(k+1)!}\right)^{2}}{\left(\dfrac{z^{k}}{k!}\right)^{2}}\right|&=\lim_{k\rightarrow\infty}\left(\dfrac{|z|}{k+1}\right)^{2}=0, \end{align*} so interval of convergence is the whole $\{|z|<\infty\}$, so the series converges on the whole plane. And we know that it is continuous on each compact box, and hence continuous on the whole plane.