If $(a_n)$ diverges, then when does it tend to infinity

Question

Let $(a_n)$ be a sequence of real numbers. Consider the two points,

• 1.$\quad$ $(a_n)$ diverges,
• 2.$\quad$ $(a_n)$ tends to plus infinity.

Using the fact that an unbounded sequence diverges, the second point implies the first one. What conditions should $(a_n)$ have to make first point to imply the second one? It seems obvious to me that $(a_n)$ has to be non-negative eventually or increasing eventually, but I do not know how to prove it.

Example: Let $(b_n)$ be a sequence of positive numbers, and define $s_N=\sum_{n=1}^{N}b_n$. If $\sum_{n=1}^{\infty}b_n$ diverges, then it must be the case that $s_N\to\infty$ as $N\to\infty$, which is reasonable. But I want to know the proof. This is why I asked a question.

Answer 1

If $(a_n)$ is (weakly) increasing, then by the monotone sequence theorem it either converges to a finite value or diverges to $+\infty$. However, there are sequences that tend to $+\infty$ that are not eventually increasing, for example

$(a_n)=(1,2,3,2,3,4,3,4,5,4,5,6,5,6,7,6,7,8,...)$.

Answer 2

"Eventually non-negative" is necessary but not sufficient.

"Eventually increasing" is not necessary and not sufficient.

"Unbounded" is necessary but not sufficient.

"Eventually increasing and unbounded" is not necessary but sufficient.

As said in comments, "larger than any number $u$ as of some $n(u)$" is the straightest requirement. This can also be phrased as "bounded below by an increasing unbounded function". (Intuitively, closer and closer to infinity.)