TITLE: Is $\lim_{n\to \infty} \frac{np_n}{\sum_{i=1}^n p_i} = 2$ true?
QUESTION [5 upvotes]: Noob here. I was playing around with primes in JavaScript and I found that if I divide the nth prime times n to the sum of primes up to n, I get closer to 2 for each n going to infinity:  $$\lim_{n\to \infty} \frac{np_n}{\sum_{i=1}^n p_i} = 2$$
Because of computer limitations I could only test this up to n = 25000, where the result was around 2.1. My question is if the result will actually get closer to 2 (but never under) or that it will go under eventually. And if there is a proof, ofcourse...

REPLY [5 votes]: By Prime Number Theorem and this we get
$$
\lim_n \,\frac{np_n}{p_1+\ldots+p_n}=\lim_n\, \frac{n\,(n \ln n)}{\frac{1}{2}n^2\ln n}=2.
$$

UPDATE: With regard to the convergence from above, the following would be correct if the upper bound on the sum were exact, but it isn't. Probably a different approach is needed.
From quantitative estimate of sum of k primes and the k-th prime we get 
$$\require\cancel
\frac{np_n}{p_1+\ldots+p_n}\ge \frac{\cancel{n^2}\left(\ln n+\ln \ln n-1\right)}{\frac{\cancel{n^2} \ln n}{2}\left(1+\frac{1}{2\ln n}+\frac{4.02}{(\ln n)^2}\right)}\overbrace{=}^{t=\ln n}2 \, \frac{t+\ln t-1}{t+1/2+4.02/t}.
$$
The rightmost expression is larger than $2$ for $t>7.604$, that is $n>e^{7.604}\approx 2006.2$, so if your calculations are correct, indeed the sequence converges to $2$ strictly from above.