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This is somewhat related to Erdős conjecture on arithmetic progressions

Is there an infinite sequence of positive integers $a_n$ s.t. $\sum_{n=1}^\infty{\frac{1}{a_n}}$ converges and $a_n$ contains arbitrary long arithmetic progressions?

If one allows negative integers a solution is $a_n=(-1)^n n$

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Yes. Take your favorite arithmetic progression $\{a_n\}$ and a very fastly growing integer sequence $\{b_n\}$. In succession, add $b_1$ to all the terms of $a_n$ after the first, add $b_2$ to all the terms after the third, ..., add $b_r$ to all the terms after the $\binom{r+1}{2}$-th. If $b_n$ is at least (roughly) $n^4$, say $b_n=((n!)!)!$, this should work. – M P Jan 19 '12 at 10:56
Actually, you can use a convergent subsequence of the harmonic sequence, where the arithmetic progression starts with something like (nlog(n))^2, the harmonic sum of the progression with n terms and difference 1 will be less than 1/n(log(n))^2. So the a_i do not have to grow very quickly. Gerhard "ask Me About System Design" Paseman, 2012.01.19 – Gerhard Paseman Jan 19 '12 at 11:09
1, 10, 11, 100, 101, 102, 1000, 1001, 1002, 1003, 10000, 10001, 10002, 10003, 10004, 100000, 100001,.... – Pietro Majer Jan 19 '12 at 13:02
@Pietro Thank you, dumb question... – joro Jan 19 '12 at 13:13

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