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
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$\begingroup$ You are right. I changed the title $\endgroup$ – C Marius Jun 3 '17 at 8:10

1$\begingroup$ 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 BakerHarmanPintz, 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 (BakerHarmanPintz, 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

1$\begingroup$ 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

1$\begingroup$ 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