approximation of the sum of prime superior to the sum of integer?

Question

We have the following : $$ \sum_{i =1}^n p_i \sim 0.5 \cdot n^2 \cdot \ln n$$ But $0.5 \cdot n^2 \cdot \ln n > 0.5 \cdot n \cdot (n+1)$ but : $\sum_{i= 1}^n i = 0.5 \cdot n \cdot (n+1)$

That's why I feel like there is something wrong with the result : $0.5 \cdot n \cdot \ln n$...

Answer 1

We have $p_i > i$

Hence, it should be expected that $$\sum_{i=1}^n p_i > \sum_{i=1}^n i$$