I am working with the following sequence involving primes $$T_{\alpha}(p_n) = p_n^{\alpha} \prod_{i=1}^{n} \left( 1 - \frac{1}{p_i^{\alpha}}\right)$$ with $\alpha \in (0,1)$. It has been shown Prime numbers property. A Merten's third theorem like sequence that $T_{\alpha}(p_n) \to 0$ as $p_n \to \infty$. Numerical experiments show that for $p_n$ large enough $T_{\alpha}(p_n)$ is monotonically decreasing. However, I cannot prove this for all $\alpha \in (0,1)$. Can someone help? If this is true, then I can prove that $p_{n+1} - p_n < \frac{p_n^{1 - \alpha}}{\alpha}$ for $\alpha \in (0,1)$ for $p_n$ large enough

  • $\begingroup$ You are right. I changed the title $\endgroup$ – C Marius Jun 3 '17 at 8:10
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    $\begingroup$ This is essentially (up to epsilon perturbations in the exponent) asking for the range of $\alpha$ for which one can prove that $p_{n+1} - p_n \ll p_n^{1-\alpha}$. Currently, this is known unconditionally for $\alpha < 0.475$ by Baker-Harman-Pintz, for $\alpha < 1/2$ on RH or Legendre's conjecture (or Andrica's conjecture), and for all $\alpha < 1$ on the Cramer conjecture (or Firoozbakht's conjecture). $\endgroup$ – Terry Tao Jun 4 '17 at 16:49
  • $\begingroup$ Thank you very much for your comment! Yes ... I know that (Baker-Harman-Pintz, at least), but I was wondering if this is a good way of attacking the problem ... that of proving the monotony of a certain sequence $\endgroup$ – C Marius Jun 4 '17 at 18:09
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    $\begingroup$ It's essentially equivalent. You've already noted one direction of the equivalence; conversely, if one can prove that $p_{n+1} - p_n = o( p_n^{1-\alpha})$, then $T_\alpha(p_n)$ will eventually be monotone. $\endgroup$ – Terry Tao Jun 4 '17 at 18:10
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    $\begingroup$ The convergence isn't that fast - in particular, it's subexponential. For your question, the more relevant asymptotics are that of the ratios $1 - T_\alpha(p_{n+1}) / T_\alpha(p_n)$, which to top order is $\frac{p_n^{1-\alpha} - \alpha(p_{n+1}-p_n)}{p_n}$. $\endgroup$ – Terry Tao Jun 5 '17 at 3:45

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