If $\int_0^\infty f(x)\,dx$ exists, then there exists some function $g$ such that $f=o(g)$ and $\int_0^\infty g(x)\mathrm\,dx$ exists

Question

If $f$ is a positive function such that $\int_0^\infty f(x)\,\mathrm dx$ converges, show that there is another function $g$ such that $\lim_{x \to \infty} \frac {g(x)} {f(x)} = \infty$ and $\int_0^\infty g(x)\,\mathrm dx$ also converges.

Answer 1

For every $x\geqslant0$, let $F(x)=\int_x^\infty f(t)\,\mathrm dt$ and $g(x)=\frac{f(x)}{\sqrt{F(x)}}$, then:

• $\int_0^\infty g(t)\,\mathrm dt=2\sqrt{F(0)}$ is finite,
• $f(x)=o(g(x))$ when $x\to\infty$.

These two assertions hold for the same reason, which is that $F(x)\to0$ when $x\to\infty$.

---

Exercise: Adapt this example to show that for every nonnegative sequence $(a_n)$ such that $\sum\limits_na_n$ converges there exists some nonnegative sequence $(b_n)$ such that $a_n=o(b_n)$ and $\sum\limits_nb_n$ converges.