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...
On
compute $|1+i|$ it is greater than one and so the terms of your series do not converge to $0$ hence the series is not convergent
On
If $n >0$ then, as $(1+i) \to \infty$ we have $(1+i)^n \to \infty$.
Therefore $$\lim_{i \to \infty}\frac{(1+i)^n}{n^2}=\infty \neq 0$$
If $n <0$ then $$0 < \frac{(1+i)^n}{n^2}<\frac{1}{n^2}$$
The series is not defined if $n=0$.
On
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%.
Is $i$ the square root of $-1$ ?In any case the ratio test works,
$$\left | \frac{(1+i)^{n+1}}{(n+1)^2} / \frac{(1+i)^{n}}{n^2} \right| =\frac{n^2}{(n+1)^2} |1+i|\rightarrow |1+i| $$ so as long as $1< |1+i| $ the series diverges.
For question 2 use Leibniz criteria. The sequence $\frac{1}{\sqrt[999]{n}}$ is decreasing and now you only have to show that
$$\sum\limits_{n=1}^k \frac{(3+4i)^n}{5^n}$$ is bounded. This can be done using the sum of a geometric series,
$$\sum\limits_{n=1}^k r^n=\frac{r^{k+1}-1}{r-1}$$
So $$\left| \frac{r^{k+1}-1}{r-1}\right| \leq \frac{|r|^{k+1}+1}{|r-1|}$$
Now in our case $r=\frac{3+4i}{5}$ so $|r|=1$ and $|r-1|=\frac{2}{\sqrt{5}}$ so in fact $$\left|\sum\limits_{n=1}^k r^n\right|\leq \sqrt{5}$$