Ratios between consecutive $k$-smooth numbers An integer is $k$-smooth if none of its prime factors is greater than $k$. I am curious about ratios between consecutive $k$-smooth numbers. Having computed all $3$-smooth numbers upto $10^{15}$, it seems that the upper bound on this ratio trends towards $1$.
Here is my hypothesis,
Let $n$, $\epsilon$, and $k$ be such that you can guarantee the existence of a $k$-smooth number in the interval $\left[n, (1 + \epsilon)n \right]$. As $n \to \infty$, then $\epsilon \to 0$ for all $k > 2$.
Is this necessarily true? How would one go about proving this?

It's trivial to show that $\epsilon = 1$ for $k = 2$, since these are simply the powers of 2.
For $k = 3$, my simplistic approach was to define two consequtive $3$-smooth numbers, $x = 2^a 3^b$ and $y = 2^{a+c} 3^{b+d}$. Here, $a, b, c, d$ are integers and $a, b \ge 0$. Then, the ratio between these is $y/x = 2^{c} 3^{d}$.
For this ratio to be greater than and arbitrarily close to $1$, you have to show that
$${m \frac{\log(2)}{\log(3)}}$$
can be arbitrarily close to an integer for some integer $m$. I am unsure how to proceed from here.
It's unclear this even helps prove the hypothesis since it doesn't deal with all $n$ or all intervals, but just some specific pair of $3$-smooth numbers that might get lucky and be close to each other.
 A: We can prove that for any $k\geq3$, the sequence of the ratio between consecutive $k$-smooth numbers tends to 1. To do this, we first prove a lemma very similar to what you had suggested.

Lemma: For every pair of positive real numbers $x$ and $y$ with $\frac{x}{y}\notin\mathbb{Q}$, for every $\varepsilon>0$, there exist non-negative integers $\alpha$, $\beta$ such that $$0<\beta y-\alpha x<\varepsilon.$$
  Proof: Consider the sequence $s_n=\{\frac{y}{x}n\}$, where the brackets denote the fractional part. By (a weak version of) the Equidistribution Theorem, we can find a term $s_\beta$ (with $\beta\geq0$) of this sequence in the interval $(0,\frac{\varepsilon}{x})$, so that $\frac{y}{x}\beta=\alpha+\delta$, for $\alpha\geq0$ an integer and $0<\delta<\frac{\varepsilon}{x}$. Therefore, $\alpha$ and $\beta$ satisfy the desired conditions. $\square$

Now, take a $\varepsilon>0$. We'll find an $N$, such that the ratio between two consecutive $k$-smooth numbers greater than $N$ is less than $1+\varepsilon$. First, for every prime $p\leq k$, let $\pi_p$ be any other prime with $\pi_p\leq k$, $\pi_p\neq p$, and let $\alpha_p$, $\beta_p$ be non-negative integers such that $$0<\beta_p\ln(\pi_p)-\alpha_p\ln(p)<\ln(1+\varepsilon)\Leftrightarrow$$ $$1<p^{-\alpha_p}\pi_p^{\beta_p}<1+\varepsilon.$$ This is possible by our lemma. Afterwards, choose $$N>\prod_{\substack{p\in\mathbb{P} \\ p\leq k}}p^{\alpha_p-1}.$$ Any $k$-smooth number $s$ exceeding $N$ must have a prime factor $p$ whose exponent is at least $\alpha_p$. Therefore, $sp^{-\alpha_p}\pi_p^{\beta_p}$ will be a $k$-smooth number greater than $s$, but less than $s(1+\varepsilon)$. This is what we wanted to prove. $\blacksquare$
