Let $(a_n)$ be a divergent sequence and $\alpha = \limsup_{n \to \infty} a_n$. Is it possible for there to exist an $N$ such that $a_n > \alpha$ for all $n>N$?
(If we are interested in $a_n > \alpha$ infinitely often, then the sequence $(a_n)$ alternating between $(1+1/n)$ for $n$ odd and $-1$ for $n$ even should work, since $\alpha=1$.)
Edit: The first version of this answer was wrong. Actually there does exist a divergent sequence with the stated property.
Hint: What I said was this:
No. If $a_n>\alpha$ for all $n>N$ then $\liminf a_n\ge\alpha=\limsup a_n$: hence $\liminf a_n=\limsup a_n$ so $(a_n)$ is convergent.
That's not right - I'm assuming something that's not given. In fact $\limsup a_n=\liminf a_n$ does not imply $(a_n)$ is convergent, for example...
(Hint for that: $\limsup a_n=\liminf a_n$ does imply that $\lim a_n$ exists, but it does not imply that $(a_n)$ is convergent. What???)