Is it true that $a_n \to \infty$ implies that $a_n$ is definitely increasing? Consider a sequence $(a_n)_n$ such that $a_n \to \infty$; is it true that $a_n$ is (at least definitely) increasing?
Intuitively I would say yes, I don't see how is possible that a sequence can go to infinity without increase (at least from a certain point on); I've tried to think to a counterexample but I couldn't find it, so I've tried to prove it but I suspect that I'm doing something wrong.
By hypothesis $a_n \to \infty$, so for all $M\in\mathbb{R}$ there exists $N_M\in\mathbb{N}$ such that $n \geq N_M \implies a_n \geq M$; now (this is the step I'm unsure about) I would like to use the fact that $M\in\mathbb{R}$ is arbitrary to let $M=a_{n-1}$, so I would get that $n \geq N_M \implies a_n \geq a_{n-1}$ and this is the definition of definitely increasing function since this is valid only for $n \geq N_M$.
However I do believe that this proof is not valid, because in the definition of limit $M$ is considered fixed while $a_{n-1}$ can vary because of its dependence on $n$; so I think I'm not allowed to choose $M=a_{n-1}$. Is my suspect correct and this proof is not valid?
Hints of are more welcome that resolutions :) thanks!
The sequence
$$a_n = n + 2 \times (-1)^n$$
tends to $+\infty$ but is not definitely increasing.
A counter-example can be constructed by taking any sequence which tends to $\infty$ and replacing every other term in a way that the pattern is one of:
decrease, increase, decrease, increase... or similar
(actually some of the increases could be decreases as well, but the point is you can build in decreases throughout the whole sequence).  The replacement terms still need to tend $\infty$ as well but there are many ways to ensure that.