Proposed proof for: if $\{s_{n}\}$ is bounded, then $\{\frac{s_{n}}{n}\}$ is convergent. I have written a proposed proof for the proposition below, but I am not entirely certain if it is valid. Could someone take a look at it, and let me know if you see any errors or steps that could use more justification?
Proposition 1: Let $\{s_{n}\}$ be a sequence of real numbers. If $\{s_{n}\}$ is bounded, prove that $\{s_n/n\}$
is convergent.
Proof. Assume that $\{s_{n}\}$ is bounded. Then, by definition, there exists
$M\in\mathbb{R},M>0$ such that
$$|s_{n}|\leq M\qquad\forall n.$$
That is, 
$$
-M\leq s_{n}\leq M\qquad\forall n.
$$
It follows that
$$
\frac{-M}{n}\leq\frac{s_{n}}{n}\leq\frac{M}{n}\qquad\forall n,$$
and so 
$$
|\frac{s_{n}}{n}|\leq\frac{M}{n}\qquad\forall n.$$
Therefore, we have 
$$
|\frac{s_{n}}{n}-0|\leq\frac{M}{n}\qquad\forall n.$$
Now, for every $\epsilon>0$ and every $M>0$, there exists an $n\in\mathbb{N}$
such that $\frac{M}{n}<\epsilon$. For, suppose not: then there exists
$\epsilon>0$ and $M>0$ such that for every $n\in\mathbb{N}$, we
have $\frac{M}{n}\geq\epsilon$. It follows that 
$n\leq\frac{M}{\epsilon}\;\forall n$.
This, however, contradicts the fact that $\mathbb{N}$ is unbounded
above. Thus, our original claim must hold. 
Furthermore, if for some
$\epsilon>0$, $M>0,$ and some $n\in\mathbb{N}$, we have 
$\frac{M}{n}<\epsilon$, then $\frac{M}{p}<\epsilon$ for every $p\in\mathbb{N}$,
where $p>n$ (should I include a proof by induction of this statement as a lemma? Or is it sufficiently obvious?).
But if it is true that for every $\epsilon>0$ and every $M>0$, there
exists an $n\in\mathbb{N}$ such that $\frac{M}{n}<\epsilon$, it
implies that we can make the expression
$|\frac{s_{n}}{n}-0|$ less than any given $\epsilon$. 
Therefore, for every $\epsilon>0$, there exists an $N\in\mathbb{N}$
such that $n>N$ implies 
$|\frac{_{s_{n}}}{n}-0|\leq\frac{M}{n}<\epsilon$.
It follows that $\{\frac{s_{n}}{n}\}$ converges, and, in particular,
that $\{\frac{s_{n}}{n}\}\rightarrow0$ as $n\rightarrow\infty$. QED
Thanks in advance!
 A: You don't need to exaggerate your proof that much. You have got that $-\frac{M}{n}\leq\frac{s_n}{n}\leq \frac{M}{n}$. Now $\{M/n\}$  converges to $0$. $\{-M/n\}$ convreges to $0$. So by Sandwich Theorem $\{s_n/n\}$ converges to $0$.   
Consult this page if you are not familiar with the Squeeze/ Sandwich Theorem:
http://en.wikipedia.org/wiki/Squeeze_theorem
A: Everything was done very clearly up to the point where you wrote that 
$$\left|\frac{s_n}{n}-0\right|\le\frac{M}{n}\quad\text{for all $n$}.\tag{1}$$
The earlier part about $-M\le s_n\le M$ was unnecessary but harmless.  
After getting to (1), you wrote too much, and the argument lost clarity.  
We will show that $\lim_{n\to\infty}\frac{s_n}{n}=0$. So we  need to show that for any $\epsilon\gt 0$, there is an integer $N$ such that if $n\gt N$, then $\left|\frac{s_n}{n}-0\right|\lt \epsilon$.
Let $\epsilon\gt 0$. Let $N=\left\lceil \frac{M}{\epsilon}\right\rceil$. By (1), if $n\gt N$ then 
$$\left|\frac{s_n}{n}-0\right|\le\frac{M}{n}\lt \frac{M}{M/\epsilon}=\epsilon.$$
A: Your proof is correct in every respect. However, you do not need to elaborate the proof so much.
You asked whether you should include a proof by induction somewhere. However, it is much simpler.
Suppose $\frac{M}{n} < \epsilon$ and $p > n$. Then $\frac{1}{p} < \frac{1}{n}$, whence $\frac{M}{p} < \frac{M}{n}$
A: I think it would suffice to say:
Let $\varepsilon>0$ be given. Then $|s_n| \leq M$ for all $n \in \mathbb{N}$. This implies that $\dfrac{|s_n|}{n} \leq \dfrac{M}{n}$ which is equivalent to $\left| \dfrac{s_n}{n}\right| \leq \dfrac{M}{n}$. Choose $N> \dfrac{M}{\varepsilon}$. Then we have that $\left|\dfrac{s_n}{n} \right| \leq \dfrac{M}{n} \leq \dfrac{M}{N} < \dfrac{M}{\left(\frac{M}{\varepsilon}\right)}=\varepsilon.$ Hence, $\left\{ \dfrac{s_n}{n}\right\} \to 0$.
