Reverse Erdős-Turan

Question

If $\{a_i\}$ is a set of positive integers which contains arbitrary long arithmetic progressions, how to show that $\sum\limits_{i=1}^{\infty}\frac{1}{a_i}=\infty$?

Answer 1

Your claim is false.

Let $A = \{a_i\} = \bigcup_{k=1}^\infty A_k$, where $$ A_k = 4^k+1,\ldots,4^k+2^k. $$ On the one hand $A_k$ is an arithmetic progression of length $2^k$, and so $A$ contains arbitrarily long arithmetic progressions. On the other hand $$ \sum_{i=1}^\infty \frac{1}{a_i} \leq \sum_{k=1}^\infty \frac{2^k}{4^k} = 1. $$