Complex infinity when proving divergence

Question

My calculus course book (Adams' Calculus) does not explain why $(-1)^n$ diverges (it just says "$(-1)^n$ simply diverges"), and I tried to see why it diverges by taking its limit as $n$ approaches infinity, also I introduced $e^{\ln( \, \cdot \, )}$... but the result I get is complex infinity because of $n\ln(-1)$. Now my question is, did they use this to "prove" that it diverges, or is my method wrong because of non-real infinity or is it enough?

Answer 1

The definition of a divergent sequence $a_n$ is that there exists $\varepsilon>0$ such that for all $x\in \mathbb R$, there is no $N$ such that $|a_n-x|<\varepsilon$ for all $n>N$. This is a formal way of saying that there is no point that the sequence stays close to after a long time. If $$a_n=(-1)^n=-1,1,-1,1,...,$$ then the only sensible point $x$ that we might suggest the sequence converges to are $-1$ and $1$. But for both of these points, the sequence will jump to the other rather than staying close by. Thus, the sequence diverges.

Looking back at the formal definition, we could simply let $\varepsilon=1.$

Answer 2

Assuming n is a continuous variable and you wish to take the limit $n \to \infty$ (please clarify your question), the behavior might be more easily seen:

let $z \equiv (-1)^n = (e^{i \pi})^n = e^{i n \pi}$.

This is not "divergent" in the sense that it goes to (complex) infinity, since $|z|$ remains unity, but does not converge to a single value. It could be divergent if n is allowed to be complex.