I need some help on the following problem: 

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> 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 $n\to \infty$, ${a\_{k+1}\over{a\_k}}\to1$. 
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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!