Let $f(x)$ be continuous from $[0, +\infty)$ to $ [0, +\infty)$, and $\int_{0}^{+\infty}f(x)dx$ diverges.

Question

Prove there exists some $a>0$ such that the series $\sum_nf(an)$ diverges.

I think it can be useful to partition $[0, +\infty)$ on $[n, n+1)$ and choose some $a_n$ for every $n$. But I can't understand how to build $a$ from $a_n$. Also I think integral Koshie trapping is useful.

Answer 1

The following answer is incorrect, see the comments below.

Let $A>0$. There exists $M>0$ such that $\int_0^M f(x) dx > 3A$. Since the integral diverges we may suppose WLOG that $f(M)>0$. Since $f$ is Riemann-integrable on $[0,M]$, there exists $\delta>0$ such that for any tagged partition of $[0,M]$ with mesh $\leq \delta$, $$\left|\sum_{i=0}^{n-1} f(t_i)(x_{i+1}-x_i)-\int_0^M f(x)dx\right|\leq A$$ Consider $a<\delta$ and the partition $(0,a,\ldots,\lfloor \frac Ma \rfloor a,M)$ tagged at $(a,\ldots,\lfloor \frac Ma \rfloor a,M) $ which has mesh $\leq a$. Thus $$\left|\sum_{n=1}^{\lfloor \frac Ma \rfloor} f(na)a + f(M)\left(M-\lfloor \frac Ma \rfloor a\right) -\int_0^M f(x)dx\right|\leq A$$ and $$\sum_{n=1}^{\lfloor \frac Ma \rfloor} f(na)a\geq 2A-af(M)$$ Choosing $a<\min( \delta, \frac{A}{f(M)})$ yields $$\sum_{n=1}^{\lfloor \frac Ma \rfloor} f(na)a\geq A$$ so the series $\sum_{n\geq 1} f(na)a$ diverges, and so does $\sum_{n\geq 1} f(na)$.