I would like to show that given $f:[0,+\infty) \to \mathbb{R}$ a continuous and non negative function so that exists $a<b$, $$\sum_{n=1}^{+\infty} \frac{1}{n}\int_{na}^{nb}f(x)dx<+\infty$$
then $$\int_0^{+\infty}f(x)dx$$ converges.
I should try to cut the interval $[an, bn]$ in an intelligent way so that, later, I can use $\sum_{n=1}^{+\infty} \frac{1}{n}\int_{na}^{nb}f(x)dx$ as an upper bound for the wanted integral and a middle term in between, but I can't think of any.
Thank you in advance
First we'll use the monotone convergence theorem to write $$\sum_{n=1}^\infty \frac1n \int_{na}^{nb}f(x)\,\mathrm{d}x = \int_0^\infty \left(\sum_{n=1}^\infty \frac1n\chi_{(na,nb]}(x)\right)f(x)\,\mathrm{d}x$$ where $\chi_A(x) \in \{0,1\}$ is $\chi_A(x) = 1 \iff x \in A$.
At this point, we'll focus on that sum. Specifically, $$\begin{align*} \sum_{n=1}^\infty \frac1n\chi_{(na,nb]}(x) &= \sum_{n=\lceil x/b\rceil}^{\lceil x/a\rceil-1} \frac1n \\ &\geq \int_{\lceil{x/b}\rceil}^{\lceil{x/a}\rceil}\frac1t\,\mathrm{d}t \\ &= \ln\left(\lceil x/a\rceil/\lceil x/b\rceil\right) \\ &\to \ln(b/a)\quad \text{as } x\to\infty. \end{align*}$$
Therefore, there is some $N$ such that $x \geq N$ implies $$\sum_{n=1}^\infty \frac1n \chi_{(na,nb]}(x) \geq \frac12\ln(b/a)$$ In context, this means $$\int_N^\infty f(x)\,\mathrm{d}x \leq \frac2{\ln(b/a)}\sum_{n=1}^\infty \frac1n \int_{na}^{nb}f(x)\,\mathrm{d}x < \infty.$$