Sequences unbounded above diverges?

Question

Prove or disprove (with counterexample) the following statement:

If $a_n > 0$ for all $n \in \mathbb{N}$, and $(a_n)_{n \in \mathbb{N}}$ is not bounded from above, then $a_n \to \infty$.

I thought the answer was true, since if the sequence is unbounded from above, and $a_n$ is greater than $0$, then it'd constantly increast to infinity. But it looks like the answer is false, and I don't really see why. What's a counterexample that refutes this statement? Thank you.

Answer 1

What about $a_{1}=1$, $a_{2}=2$, $a_{3}=1$, $a_{4}=4$, $a_{5}=1$, $a_{6}=6$, ...?

Answer 2

Consider $\;n \cdot \sin(n)\;$ for example.  See Is $\lim \space (n \sin n)= +\infty$ , or $\lim \space (n \sin n)= -\infty$ , or none of these? for divergence, and Is $n \sin n$ dense on the real line? for further insight.

Answer 3

Consider the sequence $a_n=n(1+(-1)^n)$ which oscillates unboundedly,but $a_n$ does not tend to $\infty$.