increasing numbers $x_n\to\infty$ such that $\sum_{n=1}^\infty(1-x_n/x_{n+1})<\infty$

Question

Is there a sequence of increasing numbers $x_n\to\infty$ such that $\sum_{n=1}^\infty(1-x_n/x_{n+1})<\infty$? When I try, for example, $x_n=\sum_{m=1}^n1/m$, then $1-x_n/x_{n+1}$ looks like $1/(n\log n)$ which doesn't have a convergent sum. If I make the gaps smaller than $1/n$ then the sequence doesn't increase to infinity.

Answer 1

Assume $1-x_n/x_{n+1}=a_n$ to be a summable sequence where $a_n \in (0,1)$; then $x_{n+1}=x_n/(1-a_n)$. For $x_n$ to diverge to infinity, you need $\prod (1-a_n)$ to diverge to zero, or $\log\prod(1-a_n)=\sum\log(1-a_n)$ to diverge to $-\infty$. But if $a_n\rightarrow 0$, then $\log(1-a_n)\sim -a_n$, so this is not possible.