Calculate the limit $\lim_{x \to \infty} \ \frac{1}{2}\sum\limits_{p \leq x} p \log{p}$ (here p is a prime) Calculate the limit $\lim_{x \to \infty} \ \frac{1}{2}\sum\limits_{p \leq x} p \log{p}$
(here the sum goes over all the primes less than or equal to x) using the Prime Number Theorem.
I think I've managed to show by definition that the limit is infinity but couldn't think of an elegant way of calculating it using the Prime Number Theorem. Any ideas ?
 A: $$\lim_{x\to\infty}\sum\limits_{p \leq x} p \log{p}>\lim_{x\to\infty}\sum\limits_{p \leq x}1$$ $$=\lim_{x\to\infty}\pi(x)\tag{$\pi(x)$ is prime counting func.}$$ $$=\lim_{x\to\infty}\left(\frac{\pi(x)}{\frac{x}{\log x}}\right)\left(\frac{x}{\log x}\right)=\left(\lim_{x\to\infty}\frac{\pi(x)}{\frac{x}{\log x}}\right)\left(\lim_{x\to\infty}\frac{x}{\log x}\right)$$ $$=\lim_{x\to\infty}\frac{x}{\log x}\to\infty$$
Here, $\lim_{x\to\infty}\frac{\pi(x)}{\frac{x}{\log x}}=1$ by prime number theorem.
A: I will write the sum as
$S(x) = \sum\limits_{p_n \leq x} p_n \log{p_n}$.
By the prime number theorem,
$p_n \sim n \ln n$,
so,
taking liberties in what follows,
$\begin{align}
S(x) 
&\sim \sum\limits_{p_n \leq x} n \ln n \ln{(n \ln n)}\\
&= \sum\limits_{p_n \leq x} n \ln n (\ln(n)+ \ln\ln n)\\
&\approx \sum\limits_{p_n \leq x} n \ln^2 n \\
&\approx \sum\limits_{p_n \leq x} n \ln^2 n \\
&\approx \sum\limits_{n \ln n\leq x} n \ln^2 n \\
&\approx \sum\limits_{n \leq x/\ln x} n \ln^2 n \\
&\approx \int_{1}^{x/\ln x} n \ln^2 n\ dn \\
&\approx\frac{1}{2} x^2 \ln^2 x \\
\end{align}
$
since
$(x^2 \ln^2 x)'
= x^2 (2 \ln x)(1/x) + 2x \ln^2 x
= 2x \ln^2 (x)(1+1/\ln x)
\sim 2x \ln^2 (x)
$.
My conclusion,
and this can be made rigorous
with a little work
(I think),
is that
$$ \ \frac{1}{2}\sum\limits_{p \leq x} p \log{p}
\sim \frac{1}{4}x^2 \ln^2 x
$$
