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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 $(1,\infty)$.

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2 Answers

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.

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Thanks a lot! (and good night!) –  Bombyx mori Jun 17 '12 at 7:53
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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 $r>0$.

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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
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