1
$\begingroup$

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

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

Hint:

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

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The limit is $1$, not $0$; both the numerator and the denominator have $n^6$. $\endgroup$ – mr_e_man Sep 2 '19 at 3:36
  • 1
    $\begingroup$ 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 $1/n^2$. $\endgroup$ – mr_e_man Sep 2 '19 at 3:39
  • $\begingroup$ 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? $\endgroup$ – holahola Sep 2 '19 at 3:42
  • 1
    $\begingroup$ Note that if $k=6$, then this series is approximately $\sum i^n$, which does not converge. $\endgroup$ – mr_e_man Sep 2 '19 at 3:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.