A limit involving powers and sums of primes This limit is one of three from a question that was closed for lacking focus and context. I can't speak to the source of the problem but I can include my own attempts and thoughts in an effort to resuscitate it.
Let $p_n$ denote the $n^{th}$ prime number. Claim:
$$
 \lim_{n\to\infty}\frac{\sqrt[np_n]{p_n^{p_1+\cdots+p_n}}}{\sqrt[np_n]{p_1^{p_1}\cdots p_n^{p_n}}}=e^{1/4}
$$
Taking the logarithm and simplifying, it amounts to showing
$$
\lim_{n\to\infty} \frac{1}{n p_n} \sum_{k=1}^{n} p_k \log\left(\frac{p_n}{p_k}\right)=\frac{1}{4}
$$Using the very crass approximation $p_n\approx n$ and asymptotics of the hyperfactorial function $H(n) = \prod_{k=1}^n k^k$ (here $A$ is the Glashier constant, but it doesn't matter), it works out:
$$
\lim_{n\to\infty} \frac{1}{n^2}\left(\log(n)\cdot \frac{n^2+n}{2}-\log(H(n))\right)
$$
$$
=\lim_{n\to\infty} \frac{1}{2}\log(n)  + \frac{\log(n)}{2n}- \frac{1}{n^2}\log\left(e^{-n^2/4} n^{(6n^2+6n+1)/12}(A+O(n^{-2}))\right)
$$
$$
=\frac{1}{4}+\lim_{n\to\infty} \frac{1}{2}\log(n)  + \frac{\log(n)}{2n}- \frac{6n^2+6n+1}{12n^2}\log(n)+\frac{o(n)}{n^2}= \frac{1}{4}
$$Of course, this is quite crass and it's not even clear to me that replacing $p_n$ with $n$ provides a lower bound. By the Prime Number Theorem, the first asymptotic we should be using is $p_n\approx n \log(n)$, but the calculation becomes formidable at this point and I'm not sure how many terms to include in the asymptotic. Still, I believe the claim is true and would like to see a proof.
 A: I will be using these approximation from Dusart paper :
1- $\sum \limits_{r \leq n} p_r = \frac{n^2}{2} (\ln n+\ln \ln n-\frac{3}{2}+o(1))$
2- $\sum \limits_{p \leq x} \ln p = \theta(x) $ and $|\theta(x)-x| = o(\frac{x}{\ln^3 x})$ which means that in this answer I will have  $\theta(x)= x$ without changing the result since the difference in the result is $o(1)$ and in the limits it equals $0$
3- $p_n = n(\ln n+\ln \ln n-1 +o(1))$
So $ \frac{1}{n p_n} \sum \limits_{k\leq n} p_k(\ln p_n -\ln p_k) = \frac{\ln p_n}{n p_n} \sum \limits_{k \leq n} p_k - \frac{1}{n p_n} \sum \limits_{p \leq p_n} p \ln p$
the first part, namely $ \frac{\ln p_n}{n p_n} \sum \limits_{k \leq n} p_k = \frac{\ln p_n}{n p_n} \frac{n^2}{2} (\ln n+\ln \ln n-\frac{3}{2}+o(1)) = \frac{n \ln p_n}{2 p_n} (\ln n+\ln \ln n-\frac{3}{2})$ the $o(1)$ is omitted since in the limits its equal $0$ so it will not change the answer but will make our calculation easier.
And using Abel's summation we have $ \frac{1}{n p_n} \sum \limits_{p \leq p_n} p \ln p = \frac{1}{n p_n} (p_n \theta(p_n)-\int \limits_{1}^{p_n} \theta(t)dt) = \frac{1}{n p_n} (p_n p_n -\int \limits_{1}^{p_n} t dt)  $ since for our simple calculation $\theta(x)=x$ and this will not change the answer since there difference is negligible in this case, and so $ \frac{1}{n p_n} (p_n^2 -\frac{p_n^2}{2}+\frac{1}{2}) = \frac{p_n}{2n} $
Thus the answer is the limits $ \lim \limits_{n\to \infty} \frac{n \ln p_n}{2 p_n} (\ln n+\ln \ln n-\frac{3}{2})-\frac{p_n}{2n} = \frac{1}{4}$ this part i leave for you with the use of $(3)$ for approximating prime numbers
So using known NT approximation one concludes that the value $\frac{1}{4}$ is the correct answer, I think you was looking for such assurance regardless of the hard method to prove it because you are not sure about the validity of $p_n \approx n$ in the answer you included with the post giving your high reputation but I might be wrong
A: This is ONLY A COMMENT and NOT RIGOROUS. Just some ideas.
From OEIS A007504:
$$\sum_{k=1}^n{p_k} = \frac{n^2\log{n}}{2} + O(n^2\log{\log{n}})$$
and thus:
$$\sqrt[np_n]{p_n^{p_1+\cdots+p_n}} \approx (n\log{n})^{\frac{n^2\log{n}/2}{n^2\log{n}}} = (n\log{n})^{1/2}$$
From OEIS A024450:
$$\sum_{k=1}^n{p_k}^2 = \frac{n^3\log^2{n}}{3} + O(n^3\log{n}\log{\log{n}})$$
and from the weighted AM-GM inequality (see here):
$$\sqrt[np_n]{p_1^{p_1}\cdots p_n^{p_n}} \le \left(\frac{\sum_{k=1}^n{p_k^2}}{\sum_{k=1}^n{p_k}}\right)^{\frac{\sum_{k=1}^n{p_k}}{np_n}} \approx \left(\frac{\frac{n^3log^2{n}}{3}}{\frac{n^2\log{n}}{2}}\right)^{\frac{\frac{n^2\log{n}}{2}}{n^2\log{n}}}=\left(\frac{2}{3}n\log{n}\right)^{1/2}$$
And combining numerator and denominator:
$$
 \lim_{n\to\infty}\frac{\sqrt[np_n]{p_n^{p_1+\cdots+p_n}}}{\sqrt[np_n]{p_1^{p_1}\cdots p_n^{p_n}}} \ge \sqrt{\frac{3}{2}} \approx 1.225
$$
and $\sqrt{\frac{3}{2}} < e^{1/4} \approx 1.284$.
OEIS has also the hyperprimorial:
$$\prod_{k=1}^n{p_k^{p_k}}$$
at OEIS A076265, but I think the given asymptotics might be wrong:
$$\prod_{k=1}^n{p_k^{p_k}} \approx e^{n^2\log^2{n}/2}$$
because it would give:
$$\sqrt[np_n]{p_1^{p_1}\cdots p_n^{p_n}} \approx e^{\frac{n^2\log^2{n}/2}{n^2\log{n}}}=\sqrt{n}$$
and also because numeric tests do not seem to give good results, but I may be wrong.
