$ \sum\limits_{n=1}^\infty \dfrac{(1+i)^n}{n^2}$ is divergent, and no idea about $\sum\limits_{n=1}^\infty \dfrac{(3+4i)^n}{5^n\,\sqrt[999]{n}}$ How can one see that $ \sum\limits_{n=1}^\infty \dfrac{(1+i)^n}{n^2}$ is divergent, and by which criterion? I was using a binomial theorem for $ (1+i)^n $ as $ \sum\limits_{k=0}^n \dbinom{n}{k} i^n$, but it seems to get very complicated and not clear for me.
Question 2:
What is the result in this case: $\sum\limits_{n=1}^\infty \dfrac{(3+4i)^n}{5^n\,\sqrt[999]{n}}$?
The root criterion gives me 1 in this case...

This is for your second sum:
$$|3+4i|=5$$ 
$$\arg(3+4i)=\arctan(4/3)=:\alpha \approx 1$$
$$\sum\limits_{n=1}^\infty \frac{(3+4i)^n}{5^n\,\sqrt[999]{n}} =
\sum\limits_{n=1}^\infty \frac{\exp(\alpha n i)}{\sqrt[999]{n}}$$
$$\frac{1}{\sqrt[999]{n}} \downarrow 0$$
I'm pretty much certain that this sum will converge by something like Leibniz' alternating series test, but this is not exactly an alternating series so I can't say 100%.
