sum of primes: approximate closed form?

Question

Can you find this sum? $$\sum_{\text{primes } p \le n}p.$$

I don't know how to start, let alone do this sum!

Thanks for your help.

Kind regards.

Answer 1

The sum of the first n primes $p_1+\cdots +p_n$ is estimated, using $p_n\ge n\log(n)$, as follows: $$ \sum_{i=1}^np_i\ge \sum_{i=2}^n i\log(i)\ge \int_2^n (t-1)\log(t-1)dt=(n-1)^2\frac{\log(n-1)}{2}-\frac{(n-1)^2}{4}+\frac{1}{4}.$$ Furthermore one can show, using $p_n\le n(\log(n)+\log(\log(n))+1)$ for an upper bound, that $$ \sum_{i=1}^np_i = n^2 \frac{\log(n)}{2}+O(n^2\log(\log(n))), $$ which gives you an "approximate expression" with an error term. If you write $\sum_{p\le m}p$ for your sum, then this applies to your sum.

Answer 2

The $n$-th prime number ($p_n$) is asymptotic to $n \log n$, as a consequence of the prime number theorem. So the sum of the first $K$ prime numbers is $\sum_{i=1}^{K} p_i \sim (1/2) K^2 \log K \sim (1/2) K p_K$. The sum of prime numbers no longer than $n$ is therefore $\sim (1/2) n \pi(n)\sim n^2/(2\log n)$.