# Density in $(1 , \infty)$

Let $b_n$ be a strictly monotone increasing sequence of positive real numbers which is divergent to infinity and $$\lim_{n\to\infty}\frac{b_n}{b_{n+1}}=1.$$

Show that $\left\{\frac{b_m}{b_n}\,:\,n \lt m\right\}$ is dense in $(1,\infty)$.

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Can you fix the question statement please? – copper.hat Jul 27 '12 at 16:45
I cannot type the entire question . It should be{b(m)/b(n):n<m} is dense in (1 ,oo) – Ester Jul 27 '12 at 16:52
@t.b. My apologies, but it wasn't exactly the same edit each time. I was trying to edit the title and I didn't see any preview until I submitted it. The third edit was only a space because someone else beat me to the meat of the edit I was trying to do. Please feel free to reject any edits you don't like. I for one won't be offended. – Code-Guru Jul 27 '12 at 17:14
@t.b. Didn't you copy and paste the your edits before you reloaded mine? =p – Code-Guru Jul 27 '12 at 17:25
@Sopu: It doesn't make too much sense to put a bounty on the question because "the current answers do not contain enough detail" without pointing out which details you're missing in the existing answer. – joriki Jul 29 '12 at 18:10

It may be conceptually easier to look at the sequence $a_n=\log b_n$ instead. Then the premises are that $a_n\to\infty$ but $a_n-a_{n+1}\to 0$ for $n\to \infty$, and we want to show that $D=\{a_m-a_n\mid 1<n<m\}$ is dense in $\mathbb R_+$. (Because the logarithm is a homeomorphism, it preserves limits and density in both directions).

To see that $D=\{a_m-a_n\mid 1<n<m\}$ is dense, let an open interval $(x,x+\delta)\subseteq \mathbb R_+$ be given; we then need to show that this interval contains a point in $D$.

Because $a_k-a_{k+1}\to 0$ we can choose $n$ such that $|a_m-a_{m-1}|<\delta$ for all $m\ge n$. Also, because $a_k\to\infty$ there are $m$s such that $a_m>x+a_n$. Chose the least $m>n$ with this property.

Because we have chosen $m$ to be least possible, we have $a_{m-1}\le x+a_n$, and by our choice of $n$ we then know $a_m < x+a_n+\delta$. Putting these inequalities together, we have $x+a_n < a_m < x+a_n+\delta$ or $a_m-a_n\in(x,x+\delta)$, as required.

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Since $b_n$ diverges to $\infty$, the product

$$\frac{b_{n+1}}{b_n}\frac{b_{n+2}}{b_{n+1}}\frac{b_{n+3}}{b_{n+2}}\cdots$$

must also diverge to $\infty$ for all $n$. On the other hand, for any $\rho\gt1$, there is an $n$ such that $b_{k+1}/b_k\lt\rho$ for all $k\ge n$. Thus, to get arbitrarily close to a desired value of $b_m/b_n$, we just have to go far enough out that the individual factors in the above product are small enough, then multiply them up until the product exceeds the desired value.

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If $b_n = n$, then $a_{n,m} = \frac{b_n}{b_{n+1}}\frac{b_{n+1}}{b_{n+2}}\frac{b_{n+2}}{b_{n+3}}\cdots \frac{b_{n+m-1}}{b_{n+m}} = \frac{n}{n+m}$. Thus $a_{n,\infty} = \lim_{m \to \infty} a_{n,m} = 0$, hence $\lim_{n \to \infty} a_{n,\infty} = 0$. – copper.hat Jul 27 '12 at 17:04
Sorry, I had it the wrong way around -- just fixed it. – joriki Jul 27 '12 at 17:06
I don't seem to have used the monotonicity -- don't know whether that's good or bad... :-) – joriki Jul 27 '12 at 17:06
If they are not strictly increasing, $b_m/b_n$ might not be in $(1,\infty)$. – Robert Israel Jul 27 '12 at 17:21
@joriki: Remember that even if the sequence is not strictly increase, but it diverges with the assumed ratio limit, then it has a strictly increasing subsequence, therefore the proposition would be true for the sequence itself (since it would contain a dense set). I suppose that the monotonicity is intended to assure that we only remain in $(1,\infty)$. – Asaf Karagila Jul 27 '12 at 17:36