Here is a question I have asked on Math Stack Exchange https://math.stackexchange.com/questions/2290917/prime-numbers-property-mertens-theorem-related-sequence , that I would like this community to address. Yes someone responded there, but it seems to have some errors. So: Merten's third theorem states that $$ \lim_{n \to \infty} \log(p_n) \prod_{i=1}^n \left( 1 - \frac{1}{p_i}\right) = e^{-\gamma}$$ This is a particularly interesting product since $\log(p_n) \rightarrow \infty$ and $\prod_{i=1}^n \left( 1 - \frac{1}{p_i}\right) \rightarrow 0 $ hence Merten's theorem states a sort of balance between the two sequences ... Now obviously $$ \lim_{n \to \infty} p_n \prod_{i=1}^n \left( 1 - \frac{1}{p_i}\right) = \infty$$ but let $\alpha \in [0,1]$ and let $$ T_{\alpha}(p_n) = p_n^{\alpha} \prod_{i=1}^n \left( 1 - \frac{1}{p_i^{\alpha}}\right)$$ and observe that $\lim_{n \to \infty} T_{\alpha = 0} (p_n) = 0$ and $\lim_{n \to \infty} T_{\alpha = 1} (p_n) = \infty$. My question is: is there an $\alpha^* \in (0,1)$ such that $\lim_{n \to \infty} T_{\alpha^* } (p_n) = k$ with $k \in \mathbb{R}$? According to some of my numerical calculations, this is the case for $\alpha < 4/5$ and possibly above a little. The limit is $k = 0$ in these cases. But I am of course interested in a demonstration and if there is some $\alpha$ for which the limit is not zero but finite.

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    $\begingroup$ Why always when I ask a question here on MO it begins with a down vote? :)) Can't just somebody who has a problem ask a question first (or even better provide an answer for the question )? Now the guys which might know how to answer might even not look at my question because it has NEGATIVE ranking ... ! $\endgroup$
    – C Marius
    Commented May 25, 2017 at 21:49
  • $\begingroup$ Did you considered $\alpha \in (0,1)$ ? $\endgroup$
    – C Marius
    Commented May 25, 2017 at 22:08
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    $\begingroup$ Also, @CMarius, if you have a sustained identity (as opposed to "joined today"), your questions will be taken more seriously. (No, I didn't downvote, and I rarely do.) I would imagine that the community is (reasonably!) skeptical of "joined-today" people who flit in and ask a random, uninformed question. If you have a longer-term identity people will take your questions more seriously. Not surprisingly so, considering human nature, from both ends. $\endgroup$ Commented May 25, 2017 at 22:42
  • $\begingroup$ @CMarius This is not acceptable. I have already taken ~1h to answer you on MSE. As you saw in my answer and in GHfromMO's one, your level is too low for studying in details those things. $\endgroup$
    – reuns
    Commented Jun 3, 2017 at 21:55
  • $\begingroup$ @reuns if you are user1952009 from MSE you can see that your answer there contains some errors. That is the reason that I've asked here. And please if possible keep comments about the question subject only. Nobody forces you to answer any question. But what do you know? From both these answers I've learned something :) If you want to think about something here is one of my questions: mathoverflow.net/questions/271313/… $\endgroup$
    – C Marius
    Commented Jun 4, 2017 at 14:09

1 Answer 1


I don't think this question is of research level, but let me answer it. Using that the natural logarithm is a concave function, we have $$ \log T_\alpha(p_n)=\alpha\log p_n+\sum_{i=1}^n\log\left(1-\frac{1}{p_i^\alpha}\right)<\alpha\log p_n-\sum_{i=1}^n\frac{1}{p_i^\alpha}.\tag{$*$}$$ Using the Lebesgue-Stieltjes integral and then integration by parts, the sum on the right hand side equals $$\sum_{i=1}^n\frac{1}{p_i^\alpha}=\int_{2-}^{p_n}\frac{d\pi(x)}{x^\alpha}=\frac{\pi(p_n)}{p_n^\alpha}+\alpha\int_2^{p_n}\frac{\pi(x)}{x^{\alpha+1}}\,dx.$$ Using the Prime Number Theorem and the fact that the logarithm grows slowly, the right hand side equals \begin{align*} (1+o(1))\left\{\frac{p_n^{1-\alpha}}{\log p_n}+\alpha\int_2^{p_n}\frac{dx}{x^\alpha\log x}\right\} &=(1+o(1))\left\{\frac{p_n^{1-\alpha}}{\log p_n}+\frac{\alpha}{\log p_n}\int_2^{p_n}\frac{dx}{x^\alpha}\right\}\\ &=\frac{1+o(1)}{1-\alpha}\cdot\frac{p_n^{1-\alpha}}{\log p_n}. \end{align*} That is, on the right hand side of $(*)$, the sum always grows much faster than $\log p_n$, whence $$\lim_{n\to\infty}\log T_\alpha(p_n)=-\infty.$$ This means that $T_\alpha(p_n)$ tends to zero for any $\alpha\in(0,1)$.

Actually, we don't need the Prime Number Theorem to reach this conclusion, Chebyshev's bound $\pi(x)\gg x/\log x$ suffices for the purpose. Indeed, using this bound only, we can estimate the sum on the right hand side of $(*)$ as follows, starting from the display below ($*$): $$\sum_{i=1}^n\frac{1}{p_i^\alpha}\gg_\alpha\int_2^{p_n}\frac{\pi(x)}{x^{\alpha+1}}\,dx\gg \int_2^{p_n}\frac{dx}{x^\alpha\log x}\geq\frac{1}{\log p_n}\int_2^{p_n}\frac{dx}{x^\alpha}\gg_\alpha\frac{p_n^{1-\alpha}}{\log p_n}.$$

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    $\begingroup$ Why is this even needed? Is not the question essentially "For which fixed real $\alpha \gt 0$ does log (x) behave asymptotically like x$^\alpha$"? Gerhard "Which Gives The Answer 'None'" Paseman, 2017.05.25. $\endgroup$ Commented May 26, 2017 at 0:18
  • $\begingroup$ @GHfromMO I do not know how to prove a step in your demonstration sir, hence I posted a question on Math.SE: math.stackexchange.com/questions/2297387/… Would you please be kind to have a look at it? $\endgroup$
    – C Marius
    Commented May 26, 2017 at 9:21
  • $\begingroup$ @CMarius: I added more detail, so you can delete your question at MathStackexchange. $\endgroup$
    – GH from MO
    Commented May 26, 2017 at 16:00
  • $\begingroup$ great! Thank you! $\endgroup$
    – C Marius
    Commented May 26, 2017 at 16:04

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