This is a tough one, and I have little idea on how to prove it.
Prove that the sequence defined by $\displaystyle f(n)=(-1)^n,n\in N$ is divergent.
This must be proven by contradiction.
Suppose $\displaystyle (-1)^n\xrightarrow [n\to\infty]{}a$ , for some $\displaystyle a\in\mathbb{R}\Longrightarrow \forall\,\epsilon>0\,\,\,\exists\,N_\epsilon\in\ma thbb{N}\,\,\,s.t.\,\,\,n>N_\epsilon\Longrightarrow \,|(-1)^n-a|<\epsilon$ .
Let us choose $\displaystyle \epsilon=\frac{1}{4}\Longrightarrow$ for some natural number $\displaystyle N_\frac{1}{4}$ , we have that $\displaystyle n>N_\frac{1}{4}\Longrightarrow |(-1)^n-a|<\frac{1}{4}$ $\displaystyle \Longleftrightarrow a-\frac{1}{4}<(-1)^n<a+\frac{1}{4}$ .
But this means that all but a finite number of the sequence's elements are between the left and the right ends, and the difference between these two ends is $\displaystyle a+\frac{1}{4}-(a-\frac{1}{4})=\frac{1}{2}$ , and this is a contradiction since for any two consecutive natural numbers $\displaystyle n_1,n_2>N_\frac{1}{4}\,,\,\,|(-1)^{n_1}-(-1)^{n_2}|=2$ , so it's impossible that ALL the elements of the seq. (but perhaps a finite number of them) are within two numbers whose difference is $\displaystyle 0.5$ ...
Tonio