I have been trying to solve the following problem
Assume $f : \mathbb{R} \to (0, \infty)$ is a function satisfying $\lim_{x \to 0} \frac{f(x)}{x} = 0$. Prove that there exists a sequence $(x_n) \subset \mathbb{R}$ such that the series $\sum_{n = 1}^\infty x_n$ diverges, but $\sum_{n = 1}^\infty f(x_n)$ converges.
While the claim seems pretty obvious, I feel I lack the necessary intuition to prove it.
I don't know how to prove by contradiction, but here's a proof:
For every $k$, there exists $N_k$ so that whenever $0\leq x\leq \frac{1}{N_k}$ we have $f(x) < \frac{1}{k} x$. By enlarging $N_k$ if necessary we may assume $N_{k+1}>2N_k$.
For any sequence $x_n$, let $a_k$ denote the number of $n$'s where $x_n\in (\frac{1}{2N_k},\frac{1}{N_k})$. Then
$$\sum_n f(x_n) \leq \sum_k\frac{a_k}{N_k\cdot k}$$ because $a_k$ of the $x$'s are in $(\frac{1}{2N_k},\frac{1}{N_k})$ and so are at most $\frac{1}{N_k}$, but $f(x) \leq \frac{x}{k}$.
While for similar reasons $$\sum_n x_n \geq \sum_{k} \frac{a_k}{2N_k}$$
Now we just need to choose $a_k$ properly. So we will just choose $a_k = \lfloor N_k/\sqrt{k}\rfloor$, these bracket denotes the largest integer smaller than $\frac{N_k}{\sqrt{k}}$, its a technicality because $a_k$ needs to be an integer. Namely, from any interval between $(\frac{1}{2N_k},\frac{1}{N_k})$ we will arbitrarily choose $a_k$ elements. This guarantees that
$$\sum_n x_n \geq \sum_k (\frac{1}{2\sqrt k}-\frac{1}{N_k})$$ Note that $\sum_k \frac{1}{2\sqrt{k}}$ diverges, on the other hand, since $N_{k+1}>2N_k$ we have $N_k>2^{k-1}$ which is huge, so $\frac{1}{N_k}$ is too small to change the divergence.
On the other hand $$\sum_n f(x_n) \leq \sum_k \frac{1}{k^{1.5}}$$ which conveges.
