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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 ?

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    $\begingroup$ $p$ and $\log p$ are $>1$ for sufficiently large $p$, and there are infinitely many primes, so it must diverge. $\endgroup$ – Angela Pretorius Jun 25 '13 at 14:34
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    $\begingroup$ Are you sure you don't mean $\frac{\log p}{p}$ or something similar? $\endgroup$ – A.S Jun 25 '13 at 14:35
  • $\begingroup$ @Andrew Salmon This is how the question appeared in the exam. $\endgroup$ – Robert777 Jun 25 '13 at 14:40
  • $\begingroup$ @Andrew Salmon I would love to see an elegant answer to your question as well (in case you have one) $\endgroup$ – Robert777 Jun 25 '13 at 14:41
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$$\lim_{x\to\infty}\sum\limits_{p \leq x} p \log{p}>\lim_{x\to\infty}\sum\limits_{p \leq x}1$$ $$=\lim_{x\to\infty}\pi(x)\tag{$\pi(x)$ is prime counting func.}$$ $$=\lim_{x\to\infty}\left(\frac{\pi(x)}{\frac{x}{\log x}}\right)\left(\frac{x}{\log x}\right)=\left(\lim_{x\to\infty}\frac{\pi(x)}{\frac{x}{\log x}}\right)\left(\lim_{x\to\infty}\frac{x}{\log x}\right)$$ $$=\lim_{x\to\infty}\frac{x}{\log x}\to\infty$$

Here, $\lim_{x\to\infty}\frac{\pi(x)}{\frac{x}{\log x}}=1$ by prime number theorem.

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  • $\begingroup$ I think that the first inequality should be in the opposite direction. $\endgroup$ – Robert777 Jun 25 '13 at 14:59
  • $\begingroup$ @Robert777: yeah. Thanks for pointing out. $\endgroup$ – Aang Jun 25 '13 at 15:00
  • $\begingroup$ Okay, I got it. Thanks a lot ! $\endgroup$ – Robert777 Jun 25 '13 at 15:00
  • $\begingroup$ Do you really need to pull out the prime number theorem, since $\pi(x)\to\infty$ as $x\to\infty$ simply because there are infinitely many primes? $\endgroup$ – Thomas Andrews Jun 25 '13 at 15:18
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    $\begingroup$ Not really as commented by @Angela Richardson too , but OP wanted to use it. $\endgroup$ – Aang Jun 25 '13 at 15:25
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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 $$ \ \frac{1}{2}\sum\limits_{p \leq x} p \log{p} \sim \frac{1}{4}x^2 \ln^2 x $$

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