Explain that a sequence that does not go infinity has a bounded subsequence

Question

I am a TA for real analysis course. My students were asked what does the negation of $\lim \lvert x_n \rvert = \infty$ mean.

We can explicitly write the negation of that statement :

$$\exists M >0, \forall N \in \Bbb N, \exists n>N \text{ such that } \lvert x_n \rvert < M$$ It means that $x_n$ has a bounded subsequence. However, I struggle a bit to explain it because it is nearly a definition to me. I drew a line $y=M$ and tried to be more explicit "ok, if $N=1$, then a rank bigger than $1$, lets say $3$ verifies $\lvert x_3 \rvert <M$, then if $N=2$ then a rank bigger than $2$, lets say $17$ verifies $\lvert x_{17} \rvert <M$, etc so that the subsequence $x_3, x_{17},\dots$ is bounded. Some of them understood it but some still cannot get it. What is the best explanation you can give ?

Answer 1

My best explanation: If $|x_n|$ does not tend to infinity then in some sense it has to be "bounded once in a while" (the M) and so the definition is saying no matter how big you let N get ($\forall N > \mathbb{N})$ you will always find a number in the sequence ($\exists n > N)$ that is bounded by the M.

Answer 2

I guess mentioning the word subsequence distracted some students a bit from understanding the point of the statement. Because we don't need to think of a subsequence to parse the meaning of the result.

How about this one? There always be an element preventing the sequence from being ultimately larger than some bound no matter how far we count.