$\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 $\lim \sup  _{n\to \infty} r(n,k)=\infty , \forall k \in \mathbb N$ ?
 A: 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
$$r(n,k)=p_n\,\frac{p_{n+1}}{p_{n+k+1}}\frac{p_{n+2}}{p_{n+k+1}}\dotsm \frac{p_{n+k-1}}{p_{n+k+1}}\frac{p_{n+k}}{p_{n+k+1}}>\frac{p_n}{2^{k+(k-1)+\dots+2+1}}=\frac{p_n}{2^{\tfrac{k(k+1)}{2}}}\xrightarrow[n\to\infty]{}+\infty
$$
A: $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 $1$.
A: $r(n,k) \geq \frac{p_n^{k+1}}{p_{n+k+1}^k}$ 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.
