I came up with two methods. Is one of them the one you used?
Method 1:
(I prefer this one, as it "finds" the asymptote)
Using $\displaystyle p_n \sim n \ln n$ :
$\displaystyle
\sum_{k=1}^n p_k \sim \sum_{k=1}^n n\ln n = \ln \prod_{k=1}^n k^k \sim
$
Now, using the result of
this thread:
$\displaystyle
\sim \ln C + \frac{1}{2}\left(n^2+n+\frac{1}{6}\right)\ln n - \frac{1}{4}n^2 \sim \frac{1}{2}n^2\ln n
$
Method 2:
Applying Stolz theorem
$\displaystyle
\lim_{n\rightarrow \infty}\frac{n^2 \ln n}{\sum_{k=1}^n p_k} = \lim_{n\rightarrow \infty}\frac{n^2 \ln n - (n-1)^2 \ln (n-1)}{p_n} =
$
Using $\displaystyle p_n \sim n \ln n$ :
$\displaystyle
= \lim_{n\rightarrow \infty}\frac{n^2 \ln n - (n-1)^2 \ln (n-1)}{n \ln n} = \lim_{n\rightarrow \infty}\frac{n \ln(1+\frac{1}{n-1}) + 2\ln(n-1) - \frac{\ln(n-1)}{n}}{\ln n} = 2
$
Therefore $\displaystyle \sum_{k=1}^n p_k \sim \frac{1}{2}n^2\ln n$, as desired.