Let $p_n$ denotes the $n$-th prime. There are at least $p_n- 1$ primes between $p_n$ and $\prod_{k=1}^n p_k$·

Question

I wanted to read about this proposition.

(Gentile, 1950) Let $p_n$ denote the $n$-th prime. There are at least $p_n- 1$ primes between $p_n$ and $\prod_{k=1}^n p_k$·

This is an exercise in Władysław Narkiewicz's book The Development of Prime Number Theory. There he gives the source of the exercise:

Gentile, G. (1950): Numeri primi in un intervallo particolare. Period. Mat. (4), 28, 130. MR 12 p. 243.

But I cannot find this source in the internet. I do not even know what Period. Mat. means, if it is a magazine or a publishing house. I was not able to come up with a proof by myself. My question is: Has anyone an idea about the source I provided and/or can give some clue about how to prove the statement?

We need to show that for all $n \ge 3$ $$ \pi\left(\prod_{k=1}^n p_k\right)- \pi(p_n)\geq p_n-1. $$ which is equal to $$ \pi\left(\prod_{k=1}^n p_k\right)\geq p_n+n-1. $$ So it is about a lower bound for the product of the first $n$ primes.

Answer 1

One can finish this elementarily. Note that $\prod_{r=1}^n p_r = p_n\#$.

Claim $1$: $\pi(p_n\#) \geq \pi(p_n^2)$ for $n \geq 3$.

Proof: Recall that for $n\geq 2$, $p_n\# -1$ has prime divisors none of which are in $\{p_1, p_2, \dots, p_n\}$. It follows that $\exists p_k > p_n$ such that $p_k \mid n\# - 1$. Thus $p_{n+1} \le p_k \le n\# - 1$. It follows for $n\geq 2$ that $p_{n+1} < p_n\#$.

Using this knowledge it thus follows that for $n\geq 3$, $\pi\left(\prod_{r=1}^n p_k\right) = \pi\left(p_n\color{blue}{\prod_{r=1}^{n-1} p_k}\right) \geq \pi\left(p_n\color{blue}{p_n}\right) = \pi\left(p_n^2\right)$ proving the desired claim.

Claim $2$: $\pi(n) \geq \frac{n\ln(2) - \ln(n+1)}{\ln n}$

Proof: Note that $v_p\left(\binom{n}{r}\right) = v_p(n!) - v_p((n-r)!) - v_p(r!) \le \log_p(n)$. Raising both sides by $p$ gives that $p^{v_p\left(\binom{n}{r}\right)}\le n$. Thus $$\binom{n}{r} = \prod_{p\mid n} p^{v_p\left(\binom{n}{r}\right)} \le \prod_{p \mid n} n \le n^{\pi(n)}$$

It hence follows that $\sum_{r=0}^n \binom{n}{r} = 2^n \le (n+1)n^{\pi(n)}$ and so $$\pi(n) \geq n\log_n(2) - \log_n(n+1)$$

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Hence for $n\geq 7$ ie $p_n \geq 17$ we have that $$\pi\left(p_n\#\right)\geq \pi(p_n^2) \geq \frac{p_n^2\ln\left(2\right) - \ln\left(p_n^2+1\right)}{\ln\left(p_n^2\right)}> 2p_n > p_n + n \geq p_n + n-1$$

where the third last inequality is true since for $x\geq 17$, $\frac{x^2\ln(2)-\ln(x^2+1)}{\ln(x^2)} > 2x$.

It remains to check the statement for the first $6$ primes for which a simple check yields its true that $\pi\left(p_n\#\right) \geq p_n + n -1$.

Answer 2

Note that for $n=1,2$ the statement fails, so we only consider $n\ge3$. We can check manually that it holds for $n=3$ and assume from now on that $n\ge4$. By utilizing the bound (see a short and elementary proof here What is the simplest lower bound on prime counting functions proof wise? with the log in base 2) $$\pi\left(2m\right)\ge\frac{m}{\log\left(2m\right)}$$ that holds for all $m\ge1$, it's enough to prove (letting $m:=p_{2}\cdot...\cdot p_{n}$) $$\frac{p_{2}\cdot...\cdot p_{n}}{\sum_{k=1}^{n}\log\left(p_{k}\right)}\ge p_{n}+n-1$$for all $n\ge4$. Proceed by induction. For $n=4$ we get $13.61..\ge10$. Assume validity for $n$, then for $n+1$ we need to show that $$\frac{p_{2}\cdot...\cdot p_{n+1}}{\sum_{k=1}^{n+1}\log\left(p_{k}\right)}\ge p_{n+1}+n.$$ By the induction hypothesis $$\frac{p_{2}\cdot...\cdot p_{n+1}}{\sum_{k=1}^{n+1}\log\left(p_{k}\right)}\ge\frac{\left(p_{n}+n-1\right)p_{n+1}\cdot\sum_{k=1}^{n}\log\left(p_{k}\right)}{\sum_{k=1}^{n+1}\log\left(p_{k}\right)} \ge$$$$\left(p_{n}+n-1\right)\frac{p_{n+1}}{\log\left(p_{n+1}\right)}\ge2\cdot\left(p_{n}+n-1\right)=2p_{n}+2n-2 \ge p_{n+1}+n$$ since $2p_{n}>p_{n+1}$ by Bertrand's postulate, $\frac{x}{x+\log\left(p_{n+1}\right)}\ge\frac{1}{\log\left(p_{n+1}\right)}$ holds for all $n\ge4$ and $x\ge1.4066$ and $\frac{p_{n+1}}{\log\left(p_{n+1}\right)}\ge2$ holds for all $n\ge2$.

Answer 3

Maybe a simpler solution would be to use so called "Bertrand Theorem" that is that there is at least one prime between $n$ and $2n$. That is, check the inequality for $n=2$" and then use induction plus "Bertrand Theorem".