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The prime number theorem says that the $n$th prime number is $p_n = \Theta(n \log n)$, so the series $\sum_n 1/(p_n \log p_n)$ should converge by comparison to $\sum_n 1/n (\log n)^2$. However this seems like overkill using deep mathematics. Is there a more elementary proof that $\sum_n 1/(p_n \log p_n)$ converges?

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A combinatorial proof of the bounds $$\frac{x}{\log x} \ll \pi(x)\ll \frac{x}{\log x}$$ can be found in the following two Math.StackExchange answers:

How to prove Chebyshev's result: $\sum_{p\leq n} \frac{\log p}{p} \sim\log n $ as $n\to\infty$? and Are there any Combinatoric proofs of Bertrand's postulate? Also see this answer for an application: Chebyshev: Proof $\prod \limits_{p \leq 2k}{\;} p > 2^k$

The bound is obtained by looking at the central binomial coefficient $\binom{2n}{n}$, and the primes dividing it. This approach is due to Erdos. It is not hard to see that the convergence of your series follows from these bounds.

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We need not use the prime number theorem to obtain $c_1n\log n <p_n<c_2n\log n$. There are quite elementary arguments for this, see e.g. How prove that $P_{2n}<(n+1)^2$. Then your argument should show that the series $\sum_n \frac{1}{p_n\log n}$ converges.

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  • $\begingroup$ I see in the link there are claims that $p_n < c n \log n$ for some $c > 0$, with a reference to a book, but no proofs were given in the link. If the proofs really are simple yet no one wants to reproduce them here, I'll accept this answer on faith. $\endgroup$ – user2566092 Sep 22 '13 at 18:02
  • $\begingroup$ The only results I know are things like there is a prime between $n$ and $6n/5$ for sufficiently large $n$, and then also the prime number theorem which to me is a deep result. I'll think about it and see if I can produce something besides just drawing a blank. But honestly, this claim that $p_n < c n \log n$ for some $c > 0$ seems almost equivalent to the prime number theorem. $\endgroup$ – user2566092 Sep 22 '13 at 18:11
  • $\begingroup$ @daniel Thanks, and I will think about it per your advice. I guess currently I don't understand the spectrum of results relating $p_n$ to $cn \log n$, and which ones are "easy" and which ones are hard like the strong prime number theorem. I'm ok at math, maybe I'll figure it out but if not it'd be great if someone could help me see how this problem is "easy". $\endgroup$ – user2566092 Sep 22 '13 at 18:24
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Let $k = p_n, k > 2$ and $n = \pi(k).$ Then

$$n = \pi(k) < \frac{6k}{\log k}= \frac{6p_n}{\log p_n} $$

so $$p_n > \frac{1}{6}n \log p_n > \frac{1}{6}n \log n.\tag1$$

This relies on another fact from Apostol, that

$\pi(n) < \frac{6n}{\log n}.$

Now the inequality (1) gets you the comparison in the OP. Proof of the line above is given at p. 83 of Apostol and does not rely on the prime number theorem. I wouldn't call it 'easy.'

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  • $\begingroup$ Thanks, I'll have to do some reading to see how this inequality's "difficulty to prove" compares to the proof of the prime number theorem. I'm assuming that if it comes from one page in a book, it's probably easier, which is what I was hoping for. Since this gives a direct reference to a book+page, I'll accept this soon if no one posts a self-contained proof. $\endgroup$ – user2566092 Sep 22 '13 at 19:05
  • $\begingroup$ @user2566092: You should accept Eric Naslund's answer! The proof is 3 pages or so and he seems to have found posts here. $\endgroup$ – daniel Sep 22 '13 at 19:06
  • $\begingroup$ Yep, it just went up as I was typing my comment. Thanks for your help daniel. $\endgroup$ – user2566092 Sep 22 '13 at 19:09

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