## A question about the limit of a sequence

Yesterday Prof. Feng gave a seminar which was about Furstenberg’s x2, x3 theorem. In the talk, he used the following fact:

Let $S = \{2^n3^m$: $m$, $n \in \mathbb{N}\cup\{0\}\}$. Arrange $S$ in an increasing order $\{s_1. Then

$\displaystyle\lim_{n \rightarrow \infty} \frac{s_{n+1}}{s_n} = 1$.

He said the proof was easy and it was left as an exercise, with the hint that $\frac{\log{2}}{\log{3}}$ is an irrational number.

However, after I discussed this question with some undergraduates and postgraduates, we couldn’t figure out a rigorous proof in short time as there is no good pattern in this sequence. (Yuen Chi Ho suggested that one could approach this question by using Hurwitz’s theorem, but I don’t know how to use it.)

Can anyone solve this “easy” exercise?

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### 7 Responses to A question about the limit of a sequence

1. KKK says:

ASSUME the limit $l$ exists, then it can be shown that $l=1$, as follows.
First of all, clearly $l\geq 1$. On the other hand, since $\log_3 2$ is irrational, $k \log_3 2\quad (\text{mod }1)$ is uniformly distributed on $[0,1]$. Thus for any small $\varepsilon\in(0,1)$, there is $k$ such that $m\geq k \log_3 2>m-\varepsilon$ for some natural $m$, which is equivalent to $3^\varepsilon > \frac{3^m}{2^k}\geq 1$. So if $2^k=s_p$, $\frac{s_{p+1}}{s_p}\leq 3^\varepsilon$. Since $\varepsilon$ is arbitrary, $l$ must be $1$.

2. lamwk says:

Oh, I know I should use equidristribution property somewhere, but could not work out the inequality..thanks for the “half proof”!

3. KKK says:

It turns out that we can modify the above to prove that the limit is 1 directly.
Define $t_p= \log_3 s_p \quad (\text{mod } 1)$. Let $\varepsilon>0$, choose $N$ so large such that by equidistribution (of $m\log_3 2$), for all $x\in [0,1)$, there exist $t_p\in (x,x+\varepsilon) \quad (\text{mod }1)$ for some $p. In particular, for all $n\geq N$, there exists $t_p\in (t_n,t_n+\varepsilon) \quad (\text{mod }1)$ for some $p. Suppose $\log_3 s_p= m\log_3 2+k$, then there is some natural $l$ such that $\log_3 s_n< m \log_3 2+k+l< \log_3 s_n +\varepsilon$. Letting $s_q= 2^m 3^{k+l}$, we thus have $q\geq n+1$ and $s_n< s_q. The result follows.

4. Charles Li says:

How can you guarantee that $l$ is a natural number (i.e., a non-negative integer)? Is it possible that it is negative and thus $k+l$ may be negative.

5. KKK says:

This is because $p and thus $\log_3 s_p<\log_3 s_n$.

6. Charles Li says:

Good job KKK.

Here we give an alternate proof using the theory of continued fraction. The calculations here are constructive. Let ${\alpha = \log_3 2}$. Let ${s_n = 2^a 3^b}$. Let ${\varepsilon > 0}$, then

$\displaystyle 1 \leq \frac{2^x 3^y}{2^a 3^b} < 1 + \varepsilon \ \ \ \ \ (1)$

if and only if

$\displaystyle 0 < (x-a) \alpha + (y-b) < \log(1 + \varepsilon) . \ \ \ \ \ (2)$

Suppose ${\frac{p_r}{q_r}}$ are the ${r}$-th convergent of the continued fraction of ${\alpha}$. It is known that (See Hardy, Wright, Theorem 163)

$\displaystyle \left| \alpha - \frac{p_r}{q_r} \right| \le \frac{1}{q_r^2}.$

We need a slightly stronger result, in fact by the same theorem

$\displaystyle 0 \leq q_{2m} \alpha - p_{2m} \le \frac{1}{q_{2m}}$

and

$\displaystyle 0 \leq -q_{2m+1} \alpha + p_{2m+1} \leq \frac{1}{q_{2m+1}}.$

Let ${m}$ be sufficiently large such that ${\frac{1}{q_{2m}^2} < \log(1+ \varepsilon)}$. Comparing with (2), we have both

$\displaystyle x=a+q_{2m}, y=b-p_{2m},$

and

$\displaystyle x = a-q_{2m+1}, y=b+p_{2m+1}.$

satisfy (2). However, ${x,y}$ given above may not be positive. To ensure that at least one of the set gives positive solutions, we need

$\displaystyle a > q_{2m+1} \text{ or } b > p_{2m}.$

There are only finitely many such ${a, b}$ not satisfying the above. Thus for ${n}$ sufficiently large, we can find positive integer ${x,y}$ satisfy (1).

7. lamwk says:

Thank for the nice proofs. Feel better as it seems not so easy (but somehow elementary, though).