The limit of the ratio of the elements in the set $S=\{3^n 4^m:n,m \in\Bbb N\}$ approaches to 1 I need some help on the following problem: 

Let the set $S=\{3^n 4^m:n,m \in\Bbb N\}$. Let $x_1,x_2,x_3,\dots$ be the elements in $S$ such that $x_1\lt x_2 \lt x_3\lt\dots$, then as $n\to \infty$, ${a_{k+1}\over{a_k}}\to1$. 

The following is how I approach this problem: 
Let $a_k=3^{n_1}4^{m_1}, a_{k+1}=3^{n_2}4^{m_2}\in S$, then: 
$$a_k=3^{n_1}4^{m_1}=3^{n_1}(3+1)^{m_1}=3^{n_1}\big(\sum_{i=0}^{m_1}(^{m_1}_i)3^i\big)=\sum_{i=0}^{m_1}(^{m_1}_i)3^{i+n_1},$$and, $$a_{k+1}=3^{n_2}4^{m_2}=3^{n_2}(3+1)^{m_2}=3^{n_2}\big(\sum_{i=0}^{m_2}(^{m_2}_i)3^i\big)=\sum_{i=0}^{m_2}(^{m_2}_i)3^{i+n_2}. $$
Then I could take the ratio of the two. However, I am not sure whether this approach works or not since next I will have to consider several cases and things get quite complicated. 
Any hint about the next step? 
Thanks! 
 A: The mistake is focusing on the next value in the sequence.
Just show that for every $\varepsilon > 0$, there is a $K = K_\varepsilon$, so that for all $s \in S$ with $s > K$ there is an $s' \in S$ with $1 < \dfrac{s'}{s} < e^\varepsilon$. Then this inequality is certainly fulfilled for $\dfrac{s_{k+1}}{s_k}$ if $s_k > K$.
Now, any large $s \in S$ has a large exponent for $3$ or a large exponent for $4$. So if we replace either a large power of $3$ with a "slightly" larger power of $4$ or vice versa, we reach our target.
Taking logarithms, we need to find pairs $(k,\ell)$ and $(m,n)$ of positive integers such that
$$\begin{align}
0 & < k\log 3 - \ell\log 4 < \varepsilon,\text{ and}\\
0 & < m\log 4 - n\log 3 < \varepsilon.
\end{align}$$
Since $\dfrac{\log 4}{\log 3}$ is irrational, there are solutions of these inequalities for every $\varepsilon > 0$. For example, all convergents in the continued fraction expansion satisfy
$$\left\lvert\frac{\log 4}{\log 3} - \frac{p}{q}\right\rvert < \frac{1}{q^2},$$
so all convergents with a denominator $> \dfrac{\log 3}{\varepsilon}$ satisfy
$$- \varepsilon < p\log 3 - q\log 4 < \varepsilon.$$
Since the convergents are alternatingly larger and smaller than $\dfrac{\log 4}{\log 3}$, any pair of successive convergents whose denominators are large enough gives pairs $(k,\ell)$ and $(m,n)$ as needed.
Now we can choose $K$ so that all $s \in S$ with $s > K$ have an exponent of $3$ of at least $n$ or an exponent of $4$ of at least $\ell$, that is achieved by choosing $K = 3^n\cdot 4^\ell$.
Then, for $s = 3^a\cdot 4^b > K$, choose $s' = 3^{a-n}\cdot 4^{b+m}$ if $a \geqslant n$, and $s' = 3^{a+k}\cdot 4^{b-\ell}$ otherwise.
A: Taking logs and generalizing,
I would conjecture this:
Let $a$ and $b$ be positive reals
such that
$a/b$ is irrational.
If $n$ and $m$ 
are non-negative integers
and $(s_i)_{i=0}^{\infty}$ 
is the set of elements of the form
$an+bm$  in increasing order,
then $ s_{k+1}-s_k \to 0$.
In other words,
for any $\epsilon > 0$,
there is an
$N(\epsilon)$ such that
$s_{n+1}-s_n < \epsilon$
for all $n > N(\epsilon)$.
I do not know right now how to prove this,
but I can prove that,
for any $\epsilon > 0$,
there is an $k$ such that
$s_{k+1}-s_k < \epsilon$.
Proof:
Consider the $n^2$ elements
of the form
$ai+bj$ for
$0 \le i, j \le n-1$.
The maximum value
of these is
$(n-1)(a+b)$.
If $d_k = s_{k}-s_{k-1}$
(with $s_0 = 0$),
$\sum_{k=1}^{n^2} d_k
=\sum_{k=1}^{n^2} (s_{k}-s_{k-1})
= s_{n^2}-s_0
= (n-1)(a+b)
$.
Let $D_n = \min (d_k)_{k=1}^{n^2}$.
Then
$\begin{align}
(n-1)(a+b)
&=\sum_{k=1}^{n^2} d_k\\
& \ge \sum_{k=1}^{n^2} D_n\\
& = n^2 D_n\\
\end{align}
$
so
$D_n 
\le \dfrac{(n-1)(a+b)}{n^2}
< \dfrac{a+b}{n}
$.
By choosing
$n > \dfrac{a+b}{\epsilon}$,
$D_n < \epsilon$.
