$a_n$ increasing, $\sum\frac{1}{a_n}$ diverges $\implies\exists\ n_1

Question

I want to know if the following proposition is true or false, or if it is more-or-less equivalent to Erdos conjecture.

Proposition: For any fixed $\varepsilon>0$ and any increasing real sequence $(a_n)_{n\in\mathbb{N}}$ such that $\displaystyle\sum \frac{1}{a_n}$ diverges, then $\ \exists\ n_1<n_2<n_3,$ such that $\vert (a_{n_3} -a_{n_2}) - (a_{n_2}-a_{n_1})\vert < \varepsilon. $

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Possibly related: Here, it says that, "The weaker claim that A must contain infinitely many arithmetic progressions of length 3 is a consequence of an improved bound in Roth's theorem."

Also possibly related: Has this weak version of Erdős Conjecture on arithmetic progressions been proven, or is it still an open problem?

Is it the case that if the proposition were true, then Erdos conjecture would be true?

Answer 1

The proposition is true. Indeed, if all members of the sequence $(a_n)_{n\in\mathbb N}$ are negative then it converges, which implies the required conclusion. Otherwise, replacing $(a_n)_{n\in\mathbb N}$ by its subsequence, if needed, we can suppose that $a_1>0$. Pick $\varepsilon'>0$ such that $\varepsilon'\le\varepsilon/2$ and $\varepsilon'\le a_1$ and for each natural $n$ put $f(n)=\left\lfloor \frac{a_n}{\varepsilon'}\right\rfloor$.

If for some natural $n$ we have $f(n)=f(n+2)$ then pick $n_1=n$, $n_2=n+1$, and $n_3=n+2$. Otherwise the series $\sum_{m\in f(\mathbb N)}^\infty \frac 1m$ diverges, so by the weaker claim there exist natural $n_1<n_2<n_3$ such that $(f(n_3)-f(n_2))-(f(n_2)-f(n_1))=0$. Then

$$\left|\left(\frac{a_{n_3}}{\varepsilon'}-\frac{a_{n_2}}{\varepsilon'}\right)- \left(\frac{a_{n_2}}{\varepsilon'}-\frac{a_{n_1}}{\varepsilon'}\right)\right|<2$$

and so

$$\left|(a_{n_3}-a_{n_2})-(a_{n_2}-a_{n_1})\right|<2\varepsilon'\le \varepsilon.$$