If $\{s_n\}$ is a sequence all of whose values lie inside an interval $[a, b]$ prove that $\{s_n/n\}$ is convergent. Question: If $\{s_n\}$ is a sequence all of whose values lie inside an interval $[a, b]$ prove that $\{s_n/n\}$ is convergent.
If answer it this way,is this answer correct?
Let $\varepsilon > 0$ be given. $\{S_n\}$ is contained in a bounded interval and is thus bounded; let $M$ be the maximum value of $|S_n|$ on $[a, b]$. By the Archimedean property of the real numbers, we may find a positive integer $N$ such that, for $n > N$, $1/n < \varepsilon/M$. But then, $|(S_n)/n| < |S_n|(ε/M) < ε$, so the sequence converges (to zero, in fact). (Note that there is a minor technicality here if $S_n$ is the zero sequence, but then, $S_n/n$ converges anyway.) 
It is not true that $\{S_n\}$ necessarily converges to $b$. It can converge to any point of the interval, or possibly none at all. $\{S_n/n\}$, on the other hand, will always converge to $0$, regardless of the choice of $[a, b]$. 
Edit: After thinking about it, a closed interval $[a, b]$ must contain at least one nonzero number. If you let $M = \max |x|$ on $[a, b]$ then you can get around the technicality with $S_n$ being the zero sequence, and since $\{S_n\}$ is a subset of $[a, b]$ the same inequalities follow.
 A: Your answer is correct. You are using the boundedness of the terms and the known convergence of ${1\over n}$. Since you are using the Archemedean property, as you rightly pointed out, you might need to split the case when $a=b=0$. You can always fix this by changing $M$ trivially (Like using $M+1$ as the bound). It isn't as big an issue as you pointed out.

As always, there are multiple ways of proving this.
Another very similar approach is using Squeeze Theorem. This is based on the same idea (boundedness of the terms and known convergence of ${1 \over n}$).
\begin{align}
s_n&\in[a,b]\\
\therefore a\leq s_n &\leq b\\
\mbox{Given } n\geq 0,\\
{a\over n} \leq {s_n\over n} &\leq {b \over n}\\
\lim _{n\to \infty} {1\over n} &=0\\
\lim _{n\to \infty} {a\over n} &=0\\
\lim _{n\to \infty} {b\over n} &=0\\
\therefore \lim _{n\to \infty} {s_n\over n} &=0\\
\end{align}
A: Assume $\max\{|a|,|b|\}>0$ (otherwise the sequence is null and there is not much to prove). 
Note that $|s_n|\leq M:=\max\{|a|,|b|\}$ for all $n$.
Let $\varepsilon>0$. There is $N_\varepsilon$ such that $$\frac{1}{n}<\frac{\varepsilon}{M}$$ for all $n\geq N_\varepsilon.$ 
For example, we can take $N_\varepsilon=\Bigl\lfloor\frac{M}{\varepsilon}\Bigr\rfloor +1,$ which is a natural number strictly bigger than positive real number $\frac{M}{\varepsilon}$.
Then:
$$ \left|\frac{s_n}{n}\right| < \varepsilon$$
for all $n\geq N_\varepsilon.$ 
A: Instead of taking the maximum value of $|s_n|$ over all $n$ (which may not exist in the sense that $|s_n|$ may not actually attain a maximum value), just take some positive M such that $|s_n| < M$ for all $n$. Theny you don't run into any trouble with dividing by $0$.
