Density of logarithm with fractional base

Question

Consider the set

$$C=\left\{\log_\frac{a+1}{b+1}\frac{a}{b} : a\ne b\in\mathbb{Z}^+\right\}$$ The set $C$ cannot contain all real numbers in $[1,\infty)$ because it is a countable set. But is it dense in $[1,\infty)$, or in some subinterval of it (of positive length)?

To prove that it is dense, we would need that for any $r\in[1,\infty)$ and $\epsilon > 0$, there exist $a,b$ such that $$\left|r-\log_\frac{a+1}{b+1}\frac{a}{b} \right| < \epsilon. $$ That is, we need $\log_\frac{a+1}{b+1}\frac{a}{b}$ close to $r$. The form of the expression does not allow us to solve for $a,b$ though. There was no answer when I posted on Stackexchange a while ago. I wonder what tools can be used to solve this problem.

Answer 1

$\newcommand{\ep}{\epsilon}\newcommand{\N}{\mathbb N}$The answer is no. Indeed, suppose the contrary: that the set of all values of \begin{equation*} l(a,b):=\log_\frac{a+1}{b+1}\frac ab \tag{0} \end{equation*} for distinct natural numbers $a$ and $b$ is dense in the interval $[c,d]$ for some real $c$ and $d$ such that $c<d$ and \begin{equation*} c>1. \tag{1} \end{equation*} Note that $l(b,a)=l(a,b)$. So, the set \begin{equation*} P_c:=\{(a,b)\in\N^2\colon b<a,\, l(a,b)\ge c\} \tag{2} \end{equation*} is infinite.

Lemma 1: $l(a,b)\to1$ if $1\le b<a\to\infty$.

This lemma will be proved at the end of this answer.

At this point, note that, by Lemma 1, (1), and (2), there exists some real $A_c>0$ depending only on $c$ such that $1\le b<a\le A_c$ for all $(a,b)\in P_c$. So, the set $P_c$ is finite, a contradiction.

It remains to provide

Proof of Lemma 1: Suppose that $1\le b<a<\infty$. Then \begin{equation} l(a,b)=\frac{\ln a-\ln b}{\ln(a+1)-\ln(b+1)}=\frac{\xi+1}\xi \end{equation} for some $\xi\in[b,a]$, by Cauchy's mean value theorem. So, $l(a,b)\to1$ if $b\to\infty$. On the other hand, if $b\ge1$ stays bounded while $a\to\infty$, then \begin{equation} l(a,b)\sim\frac{\ln a}{\ln(a+1)}\to1. \end{equation} This completes the proof of Lemma 1 and thus the proof of the entire claim. $\Box$