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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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 $\liminf \text{Im } a_n = -1$.


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