Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am stuck in Problem 1.3.18 in Berkeley problems in Mathematics. Without looking back in the section of solutions, I want to ask for a hint.

The problem is as follows: Let $\{b_{i}\}$ be positive real numbers with $$\lim_{n\rightarrow \infty}b_{n}=\infty$$ and $$\lim_{n\rightarrow \infty}\frac{b_{n}}{b_{n+1}}=1$$ Assume also $$b_{1}< b_{2}<b_{3}...$$ Show that the set of quotients $$\left(\frac{b_{m}}{b_{n}}\right)_{1\le n<m}$$ is dense in $(1,\infty)$.

My thoughts are to translate this problem into $a_{i}>1$, $\lim a_{i}=1$, and prove $B_{m,n}=\prod^{m}_{n}a_{i}$ is dense in $(1,\infty)$. But this does not make the problem any easier: for example, given $1+\delta$ and $\epsilon$, how can I show there must be some $m,n$ such that $B_{m,n}$ is in the $\epsilon$ neighborhood of $1+\delta$? When $a_{i}$ approach 1, it could will ignore $1+\delta$ and 'jump' straightforward from $m\in \mathbb{Z}$ to $1+\frac{1}{100}\delta$, for example. And $m\left(1+\frac{1}{100}\delta\right),m,\left(1+\frac{1}{100}\delta\right)$ may all be totally out of $1+\delta$'s $\epsilon$ neighborhood. Similarly the $a_{i}$'s after $\left(1+\frac{\delta}{100}\right)$ may shrink so quickly that both $\prod^{\infty}_{L} a_{i}$ and $m\prod^{\infty}_{L} a_{i}$ are outside of the $\epsilon$ neighborhood as well(one is too small, the other is too large). In short I do not know how to prove constructively the $B_{m,n}$s must fall into every open subset in $(1,\infty)$.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

I find it easier to think additively. Let $a_n=\ln b_{n+1}-\ln b_n$ for $n\in\Bbb Z^+$. Then your hypotheses amount to saying that $\sum_{n\ge 1}a_n$ is a divergent series of positive terms that tend to $0$, and your task is to show that $$\left\{\sum_{k=m}^na_k:1\le m<n\right\}$$ is dense in the positive reals. Here are a couple of hints:

  1. For every $m\in\Bbb Z^+$ the series $\sum\limits_{k\ge m}a_k$ diverges, so you can make $\sum\limits_{k=m}^na_k$ as big as you like.

  2. On the other hand, when $m$ is large, the partial sums of $\sum\limits_{k\ge m}a_k$ grow slowly.

share|improve this answer
Thanks a lot! (and good night!) –  Bombyx mori Jun 17 '12 at 7:53

Let $a_n = \ln b_n$, we have $\lim_{n\to\infty}a_n=\infty$, $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, and $a_1<a_2<\cdots$

First, we should build some intuition before proving rigorously. Imagine that there's a robot walks from $a_1$ then $a_2$, blahblah, eventually at $\infty$, well. The steps are really short, when $n$ is large. At every moment, say moment A, the robot can walk some consecutive steps, to approach the distance of $r$, from moment A, where $r$ is an arbitrary positive real number. There should be such a time that before that, the distance $\le r$, and after that, the distance $>r$, and the difference between the distance and $r$ is really shorter than the length of some step after moment A. Notice that the steps are really short, as the time goes. So it could be very close to $r$.

Now let's prove it rigorously. For each $r>0$ and positive integer $n$, there's one positive integer, say $\alpha_n(r)$, such that $r\ge n$, $a_{\alpha_n(r)}\le a_n+r$ and $a_{\alpha_n(r)+1}>a_n+r$, for $\lim_{m\to\infty}a_m=\infty$, therefore $$a_{\alpha_n(r)}-a_{\alpha_n(r)+1}+r<a_{\alpha_n(r)}-a_n\le r$$ Because of $\alpha_n(r)\ge n$, let $n$ tends $\infty$, we have $$\lim_{n\to\infty}\left(a_{\alpha_n(r)}-a_n\right)=r$$ so $$\lim_{n\to\infty}\frac{b_{\alpha_n(r)}}{b_n}=e^r$$ thus $e^r$ is a limit point whenever $r>0$.

share|improve this answer
OP just wanted a hint, you should edit accordingly before they see the full solution. –  Ragib Zaman Jun 17 '12 at 7:48
@RagibZaman I'm just thinking how to explain the intuition. –  Frank Science Jun 17 '12 at 7:49
I am working on other problems to avoid knowing the answer prematurely. But thanks a lot for your help. –  Bombyx mori Jun 17 '12 at 7:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.