Prove that $\sum_{k=0}^{\infty}{\frac{(3i)^k}{k!}}$ converges .

Question

by the ratio test:$$\bigg |\frac{a_{k+1}}{a_k}\bigg |=\frac{\frac{(3i)^{k+1}}{(k+1)!}}{\frac{(3i)^k}{k!}}=\frac{3i\cdot k!}{(k+1)!}=\frac{3i\cdot k!}{k!\cdot (k+1)}=\frac{3i}{k+1}\overset{k\to \infty} \longrightarrow 0$$ Is that ok?

Answer 1

Just to be cautious as I am uncertain if $i$ is positive. You might want to keep the absolute value sign.

$$\frac{3|i|}{k+1} \to 0$$