show that a given limit implies divergence 
Let $\{S_n\}$ be a sequence of positive numbers. Show that the condition $\lim_{x \to +\infty} \frac{S_{n+1}}{S_n}>1$ Implies $S_n$ goes to infinity.
Hint: Take any $r \in \mathbb{R}$ strictly between $1$ and that limit. Show that for any $n$ in natural numbers, $S_{n+1} < rS_n$ Deduce that $S_{n+2} < r^2S_n$ and $S_{n+3} <r^3S_n$

Here is what i have been thinking: So i need to bring in the definition of divergence: Let $S_n$ be a sequence of positive integers,
provided that for every number $M$ there is an integer $N$ so that
$S_n \geq M$, let $M = \frac{1}{r}$ whenever $n \geq N$. Then there exists a real number $r$ such that $r$ lies between $1$ and the limit $>1$. $S_n$ isn't bounded, and $S_n$ diverges by showing $\frac{S_{n+1}}{S_n}>1$ it isn't bounded?
 A: So philosophically what is going on here is that the ratio of the terms are (eventually) going to be greater than $1$. This means that the sequence should increase to infinity. 
Here is a proof: I think the hint is a bit off, since at the beginning of the sequence $\{s_n\}$ can be very crazy, but since $\lim s_{n+1}/s_n=p>1$, then let $r$ be a number in $(1,p)$, we can choose an $N_1$ so that $s_{n+1}/s_n>r$ whenever $n\geq N_1$. Then, $s_{n+1}>rs_n$. Note that this means: 
$$
s_{n+t}>rs_{n+t-1}>r^2s_{n+t-2}>...>r^ts_n
$$for $n\geq N_1$.
Let your $M$ be fixed. Then there is a big enough integer $N_2$ so that $M/s_{N_1}<r^{N_2}$ (this is because $r>1$) 
Let $N=N_1+N_2$. Then for $n\geq N$: 
$$
s_n>rs_{n-1}>r^2s_{n-2}>...>r^{N_2}s_{N_1}>M
$$
A: The idea is to use the fact that this ratio is, eventually, strictly greater than 1 thus, each term is times something "big" the previous term which means it is times something "big" to the $n$-th power the first term.
Specifically, since,
\begin{equation}
\lim_{n\to+\infty} \frac{S_{n+1}}{S_n}>1,
\end{equation}
there exists some $\varepsilon>0$ and $n_0\in\mathbb{N}$ such that,
\begin{equation}
\frac{S_{n+1}}{S_n}\geq 1+\varepsilon,
\end{equation}
for all $n\geq n_0$. As a result, for any $n\geq n_0$;
\begin{equation}
S_n\geq (1+\varepsilon) S_{n-1}\geq \ldots \geq (1+\varepsilon)^{n-n_0}S_{n_0}.
\end{equation}
Since $1+\varepsilon>1$ we have that $(1+\varepsilon)^n\to\infty$ as $n\to\infty$ which shows that 
\begin{equation}
\lim_{n\to\infty}S_n= \infty,
\end{equation}
i.e. $S_n$ diverges. 
