Let \(0<b_1<b_2<\cdots\) be real numbers such that \[ b_n\to\infty, \qquad \frac{b_n}{b_{n+1}}\to1. \] Prove that \[ \left\{\frac{b_m}{b_n}:1\le n<m\right\} \] is dense in \((1,\infty)\).
Solution
1. One has\[ \frac{b_{n+1}}{b_n}\longrightarrow1. \]
The hypotheses give \[ \frac{b_n}{b_{n+1}}\longrightarrow1. \] All these numbers are positive, so taking reciprocals yields \[ \frac{b_{n+1}}{b_n} = \left(\frac{b_n}{b_{n+1}}\right)^{-1} \longrightarrow1. \]
2. Fix \(x>1\) and \(\varepsilon>0\). There is \(N\) such that for every\(k\geq N\),\[ 1<\frac{b_{k+1}}{b_k} < 1+\frac{\varepsilon}{x}. \]
The lower inequality follows from the strict increase of \((b_k)\). The upper inequality holds for all sufficiently large \(k\) by step 1.
3. Fix any \(n\geq N\). There exists a least integer \(m>n\) such that\[ \frac{b_m}{b_n}\geq x. \]
Since \(b_j\to\infty\) while \(b_n\) is fixed, \[ \frac{b_j}{b_n}\longrightarrow\infty \] as \(j\to\infty\). Hence the set of integers \(j>n\) satisfying \[ \frac{b_j}{b_n}\geq x \] is nonempty and therefore has a least element \(m\).
4. For the indices \(n<m\) from step 3,\[ x \leq \frac{b_m}{b_n} < x+\varepsilon. \]
The first inequality is the definition of \(m\). By minimality, \[ \frac{b_{m-1}}{b_n}<x. \] Since \(m-1\geq n\geq N\), step 2 gives \[ \frac{b_m}{b_{m-1}} < 1+\frac{\varepsilon}{x}. \] Therefore \[ \begin{aligned} \frac{b_m}{b_n} &= \frac{b_m}{b_{m-1}} \frac{b_{m-1}}{b_n}\\ &< \left(1+\frac{\varepsilon}{x}\right)x\\ &= x+\varepsilon. \end{aligned} \]
5. The set\[ \left\{\frac{b_m}{b_n}:1\leq n<m\right\} \]is dense in \((1,\infty)\).
Given arbitrary \(x>1\) and \(\varepsilon>0\), step 4 produces an element of the displayed set in \[ [x,x+\varepsilon). \] Thus every point of \((1,\infty)\) can be approximated arbitrarily closely by elements of the set, which is exactly density.
6. Q.E.D.
Step 5 is the required conclusion.