# Function making a series convergent

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.

• To prove the existence of such a series, you should exhibit an example. Do you know about $p$-series $\sum n^{-p}$? Commented Jan 16 at 21:29
• @SammyBlack yes. I tried sequences like $\frac{1}{n}$ because the harmonic series would diverge, but $\sum_{n = 1}^\infty f(1 / n)$ not necessarily, but then a function like $f(x) = x \ln(1 + \lvert x \rvert)$ is a counterexample. The same would happen if we decreased the exponent of $p$-series further. I've also tried to prove it by contradiction, but nothing much has come out of that. Commented Jan 16 at 21:38
• not obvious. Where did you get the problem? Commented Jan 16 at 22:14
• @WillJagy it appeared in one of the tests at my uni last year. I'm practising for this year's one Commented Jan 16 at 22:18
• Does this answer your question? Functions such that $\sum \frac{1}{x_n}$ diverges $\Longrightarrow \sum \frac{1}{x_nf(x_n)}$ diverge Commented Jan 16 at 22:24

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.