Do the conditions on the following real sequence guarantee it is dense in $\ [0,1]\ ?$ I should probably not be asking any more questions because it's too late, but this one fascinated me.

If $\ (a_n)_{n\in\mathbb{N}}$ is a strictly increasing sequence of
positive real numbers such that
$$\displaystyle\lim_{n\to\infty} a_n = \infty\quad \text{and}\quad  
 \displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1,$$
then is the set:
$$ X = \left\{ \frac{a_i}{a_j}: i,j\in\mathbb{N} \right\} $$
dense in $\ [0,1]\ $ (and therefore also dense in $\ \mathbb{R}^+\ ) ?$

$$$$
Without the requirement $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1,\ $ the answer would be "no". For example, consider the sequence $\ a_n=2^n.$
So it is this requirement that makes this question interesting.
 A: The answer is yes. Here is a sketch:
Let $0< a <b < 1 $ be arbitrary.
Since $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ there exists some $N$ such that
for all $n>N$ we have
$$
\frac{a_{n+1}}{a_n} < \frac{b}{a} \,.
$$
Since $a_n$ is increasing we have
$$
\frac{a}{b}< \frac{a_{n}}{a_{n+1}} < 1 \forall n >N  \qquad (*)
$$
Next, using again $\displaystyle\lim_{n\to\infty}\frac{a_{n}}{a_{n+1}}=1$ there exists some
$M>N$ such that
$$
\frac{a_{M}}{a_{M+1}}> b 
$$
Finally, since
$$
\lim_k \frac{a_{M}}{a_{M+k}} =0 
$$
there exists some $k \geq 2$ such that
$$
\frac{a_{M}}{a_{M+k}} < b \,.
$$
Pick the minimal such $k$. Then
$$
\frac{a_{M}}{a_{M+k-1}} \geq b \\
\frac{a_{M+k-1}}{a_{M+k}} > \frac{b}{a} \qquad \mbox{ by } (*) \,.
$$
Multiplying gives $\frac{a_{M}}{a_{M+k}} >a$ which completes the proof.
P.S. Intuitivelly the solution is as follows: for $n$ large enough $\frac{a_n}{a_{n+1}}$ is arbitrary closed to $1$, but smaller than one. Multiplying consecutive terms of this form, you get telescopic products converging to $0$ is very small steps.
