$\lim_{n\to \infty} \dfrac {p_n…p_{n+k}}{p^k_{n+k+1}}$ , for fixed $k\in \mathbb N$ , where $p_n$ is the $n$-th prime

Let $p_n$ denote the $n$-th prime number . Let $r(n,k):=\dfrac {p_n...p_{n+k}}{p^k_{n+k+1}} , \forall n,k \in \mathbb N$ . I can show that for every $k \in \mathbb N , \exists n_k\in \mathbb N$ such that $r(n,k)>1 , \forall n>n_k$ .

My question is : Is it true that $\lim_{n\to \infty} r(n,k)=\infty , \forall k \in \mathbb N$ ?

If the limit doesn't exist , then is $\lim \sup _{n\to \infty} r(n,k)=\infty , \forall k \in \mathbb N$ ?

• You have k+1 terms in the numerator and k terms in the denominator. Is that what you want? – marty cohen Oct 14 '17 at 21:04

From Bertrand's postulate, we know that $$p_{n+k+1}<2p_{n+k}<2^2p_{n+k-1}<\dots <2^k p_{n+1},$$ so that $$r(n,k)=p_n\,\frac{p_{n+1}}{p_{n+k+1}}\frac{p_{n+2}}{p_{n+k+1}}\dotsm \frac{p_{n+k-1}}{p_{n+k+1}}\frac{p_{n+k}}{p_{n+k+1}}>\frac{p_n}{2^{k+(k-1)+\dots+2+1}}=\frac{p_n}{2^{\tfrac{k(k+1)}{2}}}\xrightarrow[n\to\infty]{}+\infty$$

$r(n,k) \geq \frac{p_n^{k+1}}{p_{n+k+1}^k}$ and since $k$ is fixed and $n$ is approaching $+\infty$, we can assume that $k <n$, we arrive at $r(n,k) \geq \frac{p_n^{k+1}}{p_{2n}^k} = p_n (\frac{p_n}{p_{2n}})^k$

We know that $n \ln n \leq p_n \leq 2 n \ln \frac{n}{2}$ for all $n \geq 6$.

So we get that $r(n,k) \geq p_n (\frac{n \ln n}{4n \ln n})^k = p_n \frac{1}{4^k}$

So we need to make sure that $p_n \geq 4^k$, and since $p_n \geq n$, so if $n \geq 4^k$ then $p_n \geq 4^k$ which means that $r(n,k) \geq 1$.

With the same idea $\lim \limits_{n \to \infty} r(n,k) = +\infty$, since there are infinitely many primes.

$r(n,k):=\dfrac {p_n...p_{n+k}}{p^k_{n+k+1}} , \forall n,k \in \mathbb N$

Taking logs, since $p_n \approx n\ln n$, $\ln(p_n) \approx \ln n + \ln\ln n$, so that

$\begin{array}\\ s(n,k) &=\ln(p_n...p_{n+k})-\ln(p^k_{n+k+1})\\ &=\sum_{j=0}^k \ln(p_{n+j})-k\ln(p_{n+k+1})\\ &\approx \sum_{j=0}^k (\ln(n+j)+\ln\ln(n+j))-k(\ln(n+k+1)+\ln\ln(n+k+1))\\ &= \sum_{j=0}^k (\ln n+\ln(1+j/n)+\ln(\ln n+\ln(1+j/n))\\ &\quad -k(\ln n+\ln(1+(k+1)/n)+\ln(\ln n+\ln(1+(k+1)/n))\\ &= (k+1)\ln n+\sum_{j=0}^k (\ln(1+j/n)+\ln \ln n+\ln(1+(\ln(1+j/n)/\ln n)\\ &\quad -k(\ln n+\ln(1+(k+1)/n)+\ln(\ln n+\ln(1+(k+1)/n))\\ &\approx (k+1)(\ln n+\ln\ln n)+\sum_{j=0}^k ((j/n)+(\ln(1+j/n)/\ln n))\\ &\quad -k(\ln n+\ln(1+(k+1)/n)+\ln\ln n+\ln(1+\ln(1+(k+1)/n))/\ln n)\\ &\approx \ln n+\ln\ln n+O(k^2/n)\\ &\quad -k(\ln(1+(k+1)/n)+(k+1)/(n\ln n))\\ &\approx \ln n+\ln\ln n+O(k^2/n) -k((k+1)/n)+(k+1)/(n\ln n))\\ &\approx \ln n+\ln\ln n+O(1/n) \qquad\text{since }k\text{ is fixed}\\ \end{array}$

Therefore the ratio diverges to $\infty$.

If the denominator has exponent $k+1$ (instead of $k$), the result is $O(1/n)\to 0$, so the ratio tends to $1$.