TITLE: Showing the limit is $+\infty$
QUESTION [5 upvotes]: Let $f(x)$ be continuous and positive on $[0,+\infty)$. Suppose
$$\int_0^{+\infty}\frac{1}{f(x)}dx<+\infty. $$
How can one show that
$$\lim_{s\to +\infty}\frac{\int_0^{s}f(x)dx}{s^2}=+\infty?$$

REPLY [6 votes]: For $t>0$, we have 
$$t=\int_t^{2t}\frac 1{\sqrt{f(\xi)}}\sqrt{f(\xi)}d\xi\leq \left(\int_t^{2t}\frac 1{f(\xi)}d\xi\right)^{1/2}\left(\int_t^{2t}f(\xi)d\xi\right)^{1/2}$$
so $$s^2\leq \int_s^{2s}\frac{dx}{f(x)}\cdot \int_s^{2s}f(\xi)d\xi$$
and so 
$$\frac{\int_0^{2s}f(\xi)d\xi}{s^2}\geq\frac{\int_s^{2s}f(\xi)d\xi}{s^2}\geq \frac 1{\int_s^{2s}\frac{d\xi}{f(\xi)}},$$
and we can conclude, since the fact that $\int_0^{+\infty}\frac{d\xi}{f(\xi)}$ converges implies that $\lim_{s\to \infty}\int_s^{2s}\frac{d\xi}{f(\xi)}=0$.