Something must be known about the asymptotic growth of the products of prime numbers. Let $p_n$ be the sequence of prime numbers and define $$P_k=\prod_{n=1}^k p_n$$ I'm looking for a sequence $n_k$ such that $$ \lim_{k\to\infty} \frac{P_k}{n_k} = 0 $$ where $n_k$ is as small as possible. What's the best that it is known?
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3 Answers
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We have $P_k=\exp(\vartheta(p_k))$ where $\vartheta$ is the first Chebyshev function. According to Wikipedia,
$$\vartheta(p_k)\le k\left( \log k+\log\log k-1+\frac{\log\log k-2}{\log k}\right)\text{ for }k\geq198.$$
Since we want $\vartheta(p_k)-\log n_k\to-\infty$, we can take $$\log n_k=k\left( \log k+\log\log k-1+\frac{\log\log k-2}{\log k}\right)+f(k)$$ where $f$ diverges. This gives
$$n_k=\left(\frac 1ek\log k\cdot\left(\frac{\log k}{e^2}\right)^{1/\log k}\right)^kf(k)$$ where $f$ can be anything divergent. I don't know if something better is known.
answered Dec 31, 2015 at 10:41
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$\begingroup$ Thanks, especially for being the first to point out the relation with the first Chebyschev function, whose existence was not known to me. Well, the fact that the difference goes to minus infinity does not imply that the ratio goes to zero, but I got your point. $\endgroup$ Jan 5, 2016 at 12:13
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1$\begingroup$ Oh, but it really does imply that the ratio goes to zero: $\frac{P_k}{n_k}=e^{\vartheta(p_k)-\log n_k}\to e^{-\infty}=0$ $\endgroup$ Jan 5, 2016 at 12:48
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$\begingroup$ @barto It must be $\cdots e^{f (k)}$ in the last line. $\endgroup$– user98186Jan 5, 2016 at 16:37
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$\begingroup$ @NimaBavari I know, but it doesn't really matter because $f$ is arbitrary; I thought it would be cleaner like this. $\endgroup$ Jan 5, 2016 at 18:00
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Observe that $$P_k = e^{\theta (p_k)},$$ where $\theta (x)$ is Chebyshev prime counting function. We can say of it $$\theta (p_k) < 1.000028 p_k$$ for any $k$. It follows that $$P_k = o \left (\exp (1.000028 p_k) \right),$$ which means that $$n_k = \exp (1.000028 p_k)$$ is enough.
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$\begingroup$ Thanks. How the formula with the little-o can be justified by the given upper bound? $\endgroup$ Jan 5, 2016 at 11:09
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1$\begingroup$ @user52227 Sorry for the late reply, I was at work. $$\theta (p_k) < 1.000028 p_k$$ means that for a positive quantity $m (k)$ we have $$\theta (p_k) = 1.000028 p_k - m (k).$$ Exponentiating both sides we have $$P_k = e^{-m (k)} \exp (1.000028 p_k) = o \left (\exp (1.000028 p_k) \right).$$ $\endgroup$– user98186Jan 5, 2016 at 16:20
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If we consider, instead of product of first $k$ primes, a product of all primes below $k$, then we get something called a primorial of $k$, and it's denoted $k\#$. With your notation, we have $P_k=p_k\#$. Using the prime number theorem, we can prove the following:
$$p_n\sim n\ln n,\ln(n\#)\sim n$$
Where $a_n\sim b_n$ means that $\lim\limits_{n\rightarrow\infty}\frac{a_n}{b_n}=1$. Hence we can say in a vague sense (I mean, for this, the ratio doesn't tend to $1$ like with $\sim$):
$$P_n = e^{\ln (p_n\#)} \approx e^{n \ln n} = n^n.$$
As for existence of the sequence $n_k$ you are asking for, we can use the asymptotics I've given above to prove that
$$\lim\limits_{n\rightarrow\infty}\frac{P_n}{n^{an}}=0\text{ for $a>1$}\\ \lim\limits_{n\rightarrow\infty}\frac{P_n}{n^{an}}=\infty\text{ for $a<1$}$$
And $\lim\limits_{n\rightarrow\infty}\frac{P_n}{n^{n}}$ doesn't exist (this doesn't follow from prime number theorem, but is nevertheless true). So you want the $n_k$ to be somewhat larger than $k^k$.
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$\begingroup$ Thanks. It's illuminating. I see nothing vague here: $P_n=e^{\log p_n \#} \sim e^{p_n} \sim e^{n \log n} = n^n$, so indeed $P_n \sim n^n$, or I am wrong? Having said that, one can take any divergent sequence $s_k$ and say that $\lim_{k\to\infty} \frac{P_k}{k^k} \frac{1}{s_k} =0$, so that I can take $n_k=s_k k^k$, am I doing something wrong? $\endgroup$ Jan 5, 2016 at 12:49
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2$\begingroup$ @user52227 The problem is that $f(x)\sim g(x)$ doesn't imply $e^{f(x)}\sim e^{g(x)}$. Look at $f(x)=x, g(x)=x+\log x$. $\endgroup$– WojowuJan 5, 2016 at 12:53
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$\begingroup$ Actually, $P_k = O (k^k)$. $\endgroup$– user98186Jan 5, 2016 at 16:34