Determining convergence of $\sum_{n=1}^\infty \frac{3n i^n}{(n+2i)^3}$ I've tried to solve this problem about convergence:
$\sum_{n=1}^\infty \frac{3n i^n}{(n+2i)^3}$
it's supposed to be solved using ratio, root tests or by testing the limit of the sumand. Anyways, I've tried both 3 and I had no success:
I get to a point where I'm stucked on:
$$lím_{n\rightarrow \infty} \left[\frac{n+1}{n} \left(\sqrt\frac{n^2 +4}{n^2 +2n+5}\right)^3\right]$$
Any suggestions? what would you usually do with therms like $(n+2i)^3$? I tried assuming non imaginary n values (because of the sum) and converting to polar form: $\sqrt{n^2+4}e^{i tan^{-1}(2/n)}$. 
Also I tried expanding:
$$\left|\left(\frac{n+2i}{n+1+2i}\right)^3\right| = \frac{|n+2i||n+2i||n+2i|}{|n+1+2i|^3} = \left(\sqrt\frac{n^2 +4}{n^2 +2n+5}\right)^3$$
Gelp

I just learned something new, I'd like to post it to ratify if it's true: 
$$\sum \frac{3n i^n}{(n+2i)^3} = \sum 3\frac{n i^n (n-2i)^3}{(n+2i)^3(n-2i)^3} = 3\sum \frac{ni^n (n-2i)^3}{(n^2+4)^3} $$
Expanding terms:
= $$3\sum\frac{ni^n(n^3-4n^2i-4n-2n^2i-8n+8i}{(n^2+4)^3} = 3\sum\frac{n^4 i^n-6n^3 i^{n+1} -12n^2 i^n+8n i^{n+1}}{(n^2+4)^3}$$
as every term must converge, applying the ratio test on: 
$$\sum\frac{i^n n^k}{(n^2+4)^3} \rightarrow \lim_{n\rightarrow \infty} \left|\frac{(n+1)^k}{((n+1)^2+4)^3}\right| \left| \frac{(n^2+4)^3}{n^k}\right| = \lim_{n\rightarrow \infty} \left|\frac{(n+1)^k}{n^k}\right| \left| \frac{(n^2+4)^3}{((n+1)^2+4)^3}\right|$$
First limit is 1, the second one goes to 0 (edit: to 1) because the n powers in the denominator are the same.
Is this too much going around?
