The Erdos-Turan conjecture states that if $ \{a_n\}$ is an increasing sequence of positive integers such that $ \sum_n \frac{1}{a_n} = \infty $ , then the sequence $ \{a_n\}$ contains an arithmetic progression of lenth $ k$ for every positive integer $ k$ . In 2003-2004 Ben Greene and Terence Tao proved a special case of this conjecture. In particular, they proved that the result holds for the sequence of prime numbers $ \{ p_n \}$ . One open question is whether the result holds for subsets of primes $ \{p_{n_k} \}$ such that $ \sum_k \frac{1}{p_{n_k}} = \infty$ . Below I outline an arguement which shows that we can prove the full conjecture as a consequence of this open case.

Let $ A=\{a_n\} $ be an increasing subset of positive integers such that $ \sum_n \frac{1}{a_n} = \infty$ and let $ \zeta_A (s) = \sum_k \frac{1}{{a_n}^s}$ . Then $ \zeta_A(s)$ has the an Euler product representation $ \prod_k (1-{p_{n_k}}^{-s})^{-1}$ where $ \{ p_{n_k}\}$ are the prime factors of the terms in the sequence $ \{a_n\}$ .

Taking logarithms of both sides of the Euler Product formula and expanding the logarithm, one gets $ \log \zeta_A(s) = – \sum_k \log(1 – {p_{n_k}^{-s}})$

$ = \sum_{k} \sum_{l =1}^\infty \frac{1}{{p_{n_k}}^{ls}}$ $ = \sum_{k} \frac{1}{{p_{n_k}}^s} + \sum_{k} \sum_{l=2}^\infty \frac{1}{p_{n_k}^{ls}}$ Note that we split the sum into two parts, where the first part corresponds to all terms with $ l=1$ and the second sum consists of the remaining terms. Since $ \log \zeta_A(s) \to \infty$ as $ s \to 1^+$ by hypothesis, if we show that the second sum remains bounded as $ s \to 1^+$ then we are done. So let $ R(s) = \sum_k \sum_{l=2}^\infty \frac{1}{{p_{n_k}}^{ls}}$ , then

$ |R(s)| \leq \frac{1}{2} \sum_k \sum_{l=3}^\infty \frac{1}{{p_{n_k}}^{ls}} \leq \frac{1}{2} \sum_k \frac{{p_{n_k}}^{-2s}}{1-{p_{n_k}}^{-s}} \leq \frac{1}{2} \frac{1}{1-2^{-s}} \sum_k {p_{n_k}}^{-2s} \leq \frac{1}{2} \frac{1}{1-2^{-s}} \zeta_A(2s) < \infty$ as $ s \to 1^+$ . Hence, $ \lim_{s \to 1^+} \sum_k \frac{1}{{p_{n_k}}^s} \to \infty$ . So we see that if the special open case mentioned above holds, that is if $ \{p_{n_k} \}$ contains arbitrarily long arithmetic progressions then so does the set $ A$ .

I want to know if my argument is correct and if this result has appeared in the literature.