Let $P_i$ denote the i-th prime number. Is there any formula for expressing
$$S= \sum_{i=1}^m P_i.$$
We know that there are around $\frac{P_m}{\ln(P_m)}$ prime numbers less than or equal to $P_m$. So, we have:
$$S\le m\times P_m\le \frac{P_m^2}{\ln(P_m)}.$$
I want to know, if there is a better bound for $S$, in the litrature.
Summation by parts gives $$ \begin{align} \sum_{p\le n}p &=\sum_{k=1}^n(\pi(k)-\pi(k-1))\,k\\ &=n\,\pi(n)+\sum_{k=1}^{n-1}\pi(k)(k-(k+1))\\ &=n\,\pi(n)-\sum_{k=1}^{n-1}\pi(k)\tag{1} \end{align} $$ We have that $\pi(k)=\dfrac{k}{\log(k)}\left(1+O\left(\frac1{\log(k)}\right)\right)$ and so using the Euler-Maclaurin Sum Formula, we get that $$ \sum_{k=1}^{n-1}\pi(k)=\frac12\frac{n^2}{\log(n)}+O\left(\frac{n^2}{\log(n)^2}\right)\tag{2} $$ Therefore, we get $$ \sum_{p\le n}p=\frac12\frac{n^2}{\log(n)}+O\left(\frac{n^2}{\log(n)^2}\right)\tag{3} $$ Setting $n=P_m$ should give you a closer estimate.
