How do i show that $\displaystyle\frac{\prod_{k=1}^np_k}{\sum_{k=1}^{n}p_k}$ is an integer for finitely many $n$? I have run some computations for some finitely primes to know the nature  of the ratio below (the product of the first few primes over the sum of them), specifically if it is an integer for finitely many positive integers $n$. 

My question is: How do i show thiat $\displaystyle\frac{\prod_{k=1}^np_k}{\sum_{k=1}^{n}p_k}$ is an integer for finitely many $n$'s?

Note 1: $p_1<p_2<\cdots$ is the sequence of prime numbers.
Note 2: For $n=3$ the ratio equals $3$.
 A: I did experiments, and it looks like the density of those $n$ for which the ratio is an integer approaches a limit, and the limit is approximately $0.2187.$ Of this writing I went up to $1000000$ - (with Mathematica); the convergence seems reasonably rapid, however. Note that the density is not the same as the density of square-free integers - I am not sure if there is any sort of heuristic that predicts what it should be. 
However, another experiment shows that the density of square-free numbers amongst sums of the first $n$ primes is about $6/\pi^2,$ as expected. There are two things at play here: the sum of the first $n$ primes is square free, and also its largest prime divisor is no bigger than the $n$-th prime. Now, the sum of the first $n$ primes is of order of $\frac{n^2}{2 \log n},$ and the expected value of the largest prime divisors of numbers smaller than $N$ is $\frac{\pi^2}{12} \frac{n^2}{\log n},$ not sure what the distribution looks like. Assuming the events of square-freeness and "reasonable smoothness" are independent should give a heuristic asymptotic.
UPDATE Now up to $10^7,$ and the probability of integrality is slowly declining (now $0.2174$). So, not sure what to think.
