# 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

• Try taking the modulus of the summand then taking the limit of that. That should ensure convergence – whpowell96 Sep 2 '19 at 3:18

Hint:

$$\left| \frac{3n i^n}{(n+2i)^3} \right| =\frac{3n}{\sqrt{(n^2+4)^3}}<\frac{3n}{\sqrt{(n^2)^3}} = \frac{3n}{n^3}=\frac{3}{n^2}$$

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?

• The limit is $1$, not $0$; both the numerator and the denominator have $n^6$. – mr_e_man Sep 2 '19 at 3:36
• You haven't used the fact that $k\leq4$... As suggested in the other answer, you could compare each of these series (with $k=1,2,3,4$) with $1/n^2$. – mr_e_man Sep 2 '19 at 3:39
• Oof... I really suck at this lol. Thanks for the observations. So, the limit would be 1 in this case? what about k ? as far as I understand that doesn't matter too much, right? – holahola Sep 2 '19 at 3:42
• Note that if $k=6$, then this series is approximately $\sum i^n$, which does not converge. – mr_e_man Sep 2 '19 at 3:46