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<s_2<s_3<...\}. 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?

This entry was posted in Analysis, Number Theory. Bookmark the permalink.

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<N. In particular, for all n\geq N, there exists t_p\in (t_n,t_n+\varepsilon) \quad (\text{mod }1) for some p<N. 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<s_n 3^\varepsilon. 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<n 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}}


    \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},


    \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).

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s