3
$\begingroup$

Out of couriosity and for my understanding i want to ask:

When i have the sequence $a_n = i^n$ While i is the imaginary number, i will of course have four accumulation points: $-1,1,-i,i$. So the sequence doesn't have a limit. But does it have a limes superior / inferior? My guess is no, because $\mathbb{C}$ is not a ordered field. My tutor was not able to answer me that question.

Real and imaginary part might have a lim sup/lim inf but i am not sure how to prove these.. Thanks for any hints.

$\endgroup$
  • 2
    $\begingroup$ A minor typo: *four accumulation points. $\endgroup$ – Vincenzo Oliva Dec 9 '14 at 17:36
  • $\begingroup$ hehe yeah right... $\endgroup$ – Falco Winkler Dec 9 '14 at 20:33
  • 1
    $\begingroup$ Note that $|i^n| = 1$. Thus the real and imaginary parts can't be outside of $[-1, 1]$. At the same time the real and imaginary parts take on both $-1$ and $1$ infinitely often. It should be easy to rigorously conclude what their liminfs and limsups are. $\endgroup$ – Reinstate Monica Dec 9 '14 at 20:39
  • $\begingroup$ so -1 and +1 are the lim inf and lim sup of both the imaginary part and the real part? $\endgroup$ – Falco Winkler Dec 9 '14 at 20:46
1
$\begingroup$

Yes, you need a partial ordered set to make sense of Suprema and Infima. You need this for defining $\limsup$ and $\liminf$.

If you consider the real part note that $$\text{Re } a_n = \text{Re } i^n = \begin{cases}0 &\text{ if } n \text{ odd} \\ (-1)^{\frac n2 } &\text{ if } n \text{ even}\end{cases} $$.

Then it is easy to see that $\limsup \text{Re } a_n = 1$ and $\liminf \text{Re } a_n = -1$.

Analougously one gets $\limsup \text{Im } a_n = 1$ and $\liminf \text{Im } a_n = -1$.

$\endgroup$

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.