Prove: if $\lim \frac{s_n}n=L\neq0$, then the sequence $s_n$ is not bounded.

Question

If $$\lim_{n\rightarrow \infty} \frac{s_n}{n}= L \ne 0$$ then $\{ s_n \}$ is not bounded.

How to prove this generally (i.e. without help of examples)?

I've sketched a proof below. Is it right. If not what is the flaw?

My attempt of proof:

If $$\lim_{n\rightarrow \infty} \frac{s_n}{n}= L \ne 0$$ then$ |\frac{s_n}{n}- L |< \epsilon , (n\ge N)$

$L- \epsilon < \frac{s_n}{n}< L+ \epsilon$

$\frac{s_n}{n}< L+ \epsilon \implies s_n <n (L+\epsilon)$

When $n \rightarrow \infty , s_n < \infty$

Then we can't find an N such that for any $M>0$, $|s_n |\le M$ i.e. for any $(n\ge N)$ we can't find an N such that $s_n \le -M$.

Answer 1

Suppose that $\{s_n\}$ is bounded. Then for all $n$ there exists $M>0$ such that $|s_n|\leq M$. Then we have $$-M\leq s_n\leq M ~~\forall n$$

So

$$\lim_{n\to\infty}-M/n\leq \lim_{n\to\infty} s_n/n\leq\lim_{n\to\infty} M/n.$$

By squeeze theorem, we have a contradiction:)

Answer 2

Let's continue with OP's steps. Without loss of generality, assume $L>0$, and choose $\epsilon=\frac L2$,

$$\epsilon=\frac L2, ~\exists N_1, \forall n>N_1\Longrightarrow-\frac L2=-\epsilon<\frac{s_n}n-L\Longrightarrow s_n>\frac{nL}2$$

$$\forall M>0, \exists N=\max\left(N_1, \lceil\frac{2M}L\rceil\right), \forall n>N\Longrightarrow s_n>\frac{NL}2\ge M$$

Hence, $\lim s_n\to \infty$