# Reverse Erdős-Turan

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

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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 $$\sum_{i=1}^\infty \frac{1}{a_i} \leq \sum_{k=1}^\infty \frac{2^k}{4^k} = 1.$$