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I don't think this is a valid proof, particularly the use of O notation at the end, or the series manipulations,

From a non mathematical point of view I don't understand how a proof of the prime number theorem can be given in about a paragraph as is done here and yet others go at great length to find "elementary proofs", if I am wrong about this proof can someone correct me, I only ask because I thought 'proof wiki' was a some what reliable source.

http://www.proofwiki.org/wiki/Prime_Number_Theorem

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Personally I find proof wiki to be a horrible resource for proofs (at least for group theory) as they often end up being cyclic, so I wouldn't say its always a reliable source. –  Andrew D Jan 8 '13 at 14:33
    
The proof given calls on a number of preliminary results, and is surely longer than a paragraph. –  Mark Bennet Jan 8 '13 at 14:39
    
Alot of preliminary results? like what the O term in the summation formula for the divisor function or algebraic manipulation of dirichlet series? Im sorry if your not already familiar with these results, you probably shouldn't be looking at proofs of the pnt in general. –  Ethan Jan 8 '13 at 15:05
    
Like: knowledge of the inverse and square of ζ, and knowledge of its derivative; Using a convenient property of the Moebius function; Order of Divisor Function; and by Order of Moebius Function - these may be elementary, but then so are the properties of the Mangoldt function in the same context. My point was that the length of a proof does depend on what you put in it, and what you assume. –  Mark Bennet Jan 8 '13 at 15:20
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The proof has a health warning on it. It is not analytic, and the final statement in the article is that $\displaystyle \lim_{N \to \infty} \left({ O \left({ \frac{-1}{\sqrt{N}}}\right) o \left({N}\right) - \frac{2\gamma}{N} }\right) = 0$ which does not look right since there are orders between $\sqrt N$ and $N$. –  Mark Bennet Jan 8 '13 at 15:24
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3 Answers 3

up vote 3 down vote accepted

There are many woeful technical errors, especially towards the end of this proof, but the overall strategy looks sound and can be repaired. Honestly, it feels like someone wrote down a sketch and then invited random first-year students to fill in the details with wishful thinking. I think it hardly speaks well for ProofWiki.

One of the principal ingredients is $M(n) = o(n)$, which is itself equivalent to PNT. The proof of this that ProofWiki provides is egregious. It makes the baffling claim that $$\text{“Clearly}\displaystyle \sum_{n \le N} \frac{\mu(n)}{n} \ge \sum_{n\le N} \frac{\mu(n)}{N},”$$ which is not clear to any sane reader. It's actually false at around $N = 18500$ or so.

Nevertheless, I expect one can deduce this by doing a legitimate partial summation from the convergence of $\sum \mu(n)/n$, which is proved on a separate page (this part does at least imitate the complex analysis content of Newman's proof, but I haven't looked at the details).

With this powerful result in hand, it isn't that hard to obtain PNT by elementary methods (Dirichlet hyperbola). One should be careful to choose a cutoff adapted to the implied rate of decay of $o(n)$, but I think it does go through. It just happens that the ProofWiki argument relies on enough typographic miracles that it's "not even wrong": it strikes me as little better than the old saw $$\displaystyle\frac{\sin(x)}{n} = 6.$$

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I appreciate the time you spent looking at this, I don't have very many technical skills when it comes to computers, so I was only able to get someone to mark the post as "needs explaining". –  Ethan Jan 13 '13 at 4:40
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I haven't checked the details in that page, but it says that the proof is a variant of Donald Newman's 4-page proof. It is short and elementary, yes.

Simple Analytic Proof of the Prime Number Theorem, The American Mathematical Monthly, Vol. 87, No. 9 (Nov., 1980), pp. 693-696 URL: http://www.jstor.org/stable/2321853

See also Zagier's

Newman's Short Proof of the Prime Number Theorem, The American Mathematical Monthly, Vol. 104, No. 8 (Oct., 1997), pp. 705-708 URL: http://www.jstor.org/stable/2975232

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It's worth noting that elementary is being used informally to mean "relatively easy compared to other proofs" not formally to mean "doesn't use complex analysis." The latter usage is common when discussing the PNT, as in Selberg's elementary proof of the PNT (which is much harder than Newman's proof!). –  Noah Snyder Jan 8 '13 at 14:53
    
This proof doesn't use complex analysis? Did you guys even look at the link? I am not convinced of its validity –  Ethan Jan 8 '13 at 15:03
    
@Ethan you misread the comment. Noah is saying that normally we use the term "elementary" to mean we don't use complex analysis, so lhf's use of "elementary" is non-standard... –  Thomas Andrews Jan 10 '13 at 5:22
    
@lhf I fail to see any resemblance between Newman's proof and the one on ProofWiki, which seems to be riddled with flaws. –  Erick Wong Jan 12 '13 at 3:29
    
@ErickWong, like I said, I haven't checked the proof on ProofWiki. –  lhf Jan 12 '13 at 11:54
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All the sections before "The Proof of the Prime Number Theorem" seem valid, they are very standard lemmas.

But then they start using the following identity: $$\displaystyle \frac{1}{\zeta(z)} \left({ \zeta'(z) - \zeta^2 (z) }\right) = \left({ \sum_{n \mathop = 1}^\infty \frac{\mu(n)}{n^z} }\right) \left({ \left({ \sum_{n \mathop = 1}^\infty \frac{\log(n)}{n^z} }\right) - \left({ \sum_{n \mathop = 1}^\infty \frac{d(n)}{n^z} }\right) }\right)$$ which has a sign error! The derivative of the zeta functions should be negated: $$\displaystyle \frac{1}{\zeta(z)} \left({ \zeta'(z) - \zeta^2 (z) }\right) = \left({ \sum_{n \mathop = 1}^\infty \frac{\mu(n)}{n^z} }\right) \left({ -\left({ \sum_{n \mathop = 1}^\infty \frac{\log(n)}{n^z} }\right) - \left({ \sum_{n \mathop = 1}^\infty \frac{d(n)}{n^z} }\right) }\right)$$ They make use of it with the wrong sign afterwards.

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I think the sign error crept in much earlier. They say that the Dirichlet series for $\Lambda(n)$ is $\zeta'/\zeta$ rather than $-\zeta'/\zeta$. I suspect it should be $\zeta' + \zeta^2$ they should be considering. –  Erick Wong Jan 13 '13 at 5:55
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