Divergence of a series

Question

I'd like to know if this proposition is true:

$$ \left\{ x_n \right\} | x_{n+1}>x_n \wedge \lim_{n \to \infty}x_n = \infty \Rightarrow $$ $$\sum_{n \geq 1} \left( 1 - \frac{x_n}{x_{n+1} } \right) = \infty$$ Can anyone help me?

Answer 1

Yes, this is true. You have $$ \sum_{k=1}^n \ln\frac{x_{k+1}}{x_k} = \ln x_{n+1} - \ln x_1 \to \infty. $$ If $(x_{k+1}/x_k)$ does not converge to $1$, then the series trivially diverges because the terms do not converge to $0$. Otherwise you have $x_{k+1}/x_k \le 2$ eventually, and so $$ \ln \frac{x_{k+1}}{x_k} \le \frac{x_{k+1}}{x_k}-1 \le 2 \left(1 - \frac{x_k}{x_{k+1}}\right), $$ implying your claim by comparison of series.