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 $(1,\infty)$.
 A: 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.
A: 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.
