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With $p_i$ being the $i$-th prime, I'm wondering whether there is a tighter bound than $\alpha = 4$ in the relation $$ \prod_{i=1}^n p_i < \alpha^{p_n} $$ $\alpha = 4$, which is tight enough to be used to prove Bertrand's postulate (Chebyshev's theorem) that there is always a prime in the interval $[k,2k]$, is easily proven using $\binom{2r}{r} < 4^r$ and the fact that any prime in the interval $[n/2,n]$ will divide $\binom{2r}{r} < 4^r$; see the answer to

https://math.stackexchange.com/questions/1924453/show-that-p-1p-2-cdots-p-t4n

However, that argument makes a couple of lavishly conservative statements, two of which are that since $p_n$ divides $\binom{2n}{n}$ it must be less than $\binom{2n}{n}$, and that $\binom{2n}{n} < \sum_k \binom{2n}{k}$. One might suspect that a smaller value of $\alpha$ could suffice.

Numerical experimentation finds that for $n \leq 100$,

$$ \prod_{i=1}^n p_i < (2.62)^{p_n} $$ but if you extend to $n=1000$, you need to increase to $\alpha = 2.6938$.

Thus my question: Is it known that no $\alpha < 4$ works in the inequality as $n\to \infty$, has there been a lower value demonstrated, or is this question completely open?

I would guess that this can be attacked using the Prime Number Theorem, but fluctuations from that distribution become important for this bound.

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    $\begingroup$ Look up Chebyshev functions ($\sum_p \log p$ for $p$ running over primes less than $x$). You will find $e$ is a near miss but provably an asymptotic value for your $\alpha$. Gerhard "For Some Value Of 'Provably'" Paseman, 2016.09.19. $\endgroup$ – Gerhard Paseman Sep 19 '16 at 21:32
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    $\begingroup$ While your question regarding this aspect of prime numbers is fairly basic and well known to students of analytic number theory (and so this forum is not as good a place for your question as is math.stackexchange), something that may not be known is the following, which would be a good question for this forum: If $\alpha^{p_n}$ is replaced by $\alpha^{p_{n+1}}$ in your display above, is $e$ a strict upper bound for $\alpha$ for all $n$? Gerhard "Likes This Version For MathOverflow" Paseman, 2016.09.19. $\endgroup$ – Gerhard Paseman Sep 19 '16 at 21:44
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    $\begingroup$ Bertrand, not Bertram. $\endgroup$ – abx Sep 19 '16 at 21:57
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It follows from the Prime Number Theorem that the $p_n$-th root of the product tends to $e=2.7182\dots$. In practice one takes the logarithm of the product and divides by $p_n$, which then tends to $1$. This means that any $\alpha>e$ produces an upper bound for sufficiently large $n$, while any $\alpha<e$ produces a lower bound for sufficiently large $n$.

For good practical bounds see Rosser-Schoenfeld: Approximate formulas for some functions of prime numbers. See especially (3.15) and (3.16) there, which state in particular $$1-\frac{1}{\log x}<\frac{1}{x}\sum_{p\leq x}\log p<1+\frac{1}{2\log x},\qquad x\geq 41.$$

Added. See also the recent paper of Platt-Trudgian regarding the oscillation of the average in the middle around $1$.

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For large $n$ this product behaves like $(e+o(1))^{p_n}$, this is equivalent form of Prime Number Theorem. I do not know, however, whether $e^{p_n}$ is always an upper bound.

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    $\begingroup$ It is not always an upper bound due to known oscillations of $\theta(x)-x$ of size about $\sqrt{x}$. $\endgroup$ – GH from MO Sep 19 '16 at 21:35
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    $\begingroup$ @GHfromMO if so, maybe explicit bounds combined with known values allow to find the answer to OP's question of max $\theta(x)/x$? $\endgroup$ – Fedor Petrov Sep 19 '16 at 21:39
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    $\begingroup$ Well, this is a subtle problem. It is known that $\theta(x)<x$ for $x<1.39\times 10^{17}$, and it is also known that $\theta(x)>x$ for some $x<e^{728}$. See ams.org/journals/mcom/2016-85-299/S0025-5718-2015-03021-X $\endgroup$ – GH from MO Sep 19 '16 at 21:44
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    $\begingroup$ From the last comment of @GH form MO and the right half of the inequality in his answer, we can see that $\alpha = 2.753$ is a strict upper bound, and that $\alpha = e$ is a strict lower bound. $\endgroup$ – Mark Fischler Sep 20 '16 at 15:54

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