On the asymptotic growth of the products of prime numbers

Question

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?

Answer 1

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.

Answer 2

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 $k^k$.

Answer 3

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.