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The section on Euler's proof mentions at the end that the "equation" seems odd to modern eyes, et cetera. I've seen this kind of statement in regards to Euler and other less rigorous mathematicians before, and it always annoys me. The meaning of the "equation" is perfectly evident to me; of course we'll never know for sure exactly what he meant, but how could he have meant anything other than our modern day, rigorous equivalent? The article then says that the proof can be slightly altered to conform to modern standards of rigor. Wouldn't that just amount to replacing all the infinites with Xs and taking the limit as X approaches infinity? This is almost the only useful interpretation of the infinity symbol, and the presence of the unqualified symbol naturally suggests replacing it with a limit. Why are proofs like this considered so drastically different than our modern ones? --Monguin61 22:23, 15 December 2005 (UTC)

Wouldn't that just amount to replacing all the infinites with Xs and taking the limit as X approaches infinity?

Not in this, involving "ln(ln(+∞))". The limit would merely be ∞, but what the paragraph says Euler almost certainly meant is something other than that.

This is almost the only useful interpretation of the infinity symbol

That is nonsense. There are many interpretations that are useful. You've just mentioned one of them, and what the article says Euler almost certainly meant is another one.

Why are proofs like this considered so drastically different than our modern ones?

Did you sleep through the whole 19th century? Notice, for example, the reasons why we distingish between pointwise convergence and uniform convergence, etc. Michael Hardy 00:11, 16 December 2005 (UTC)

I did write the fourth theorem.However, I'm not really sure if

${\displaystyle \sum _{n=6}^{\infty }{\frac {1}{n\ln n+n\ln \ln n\quad }}}$

really diveges. Can someone show it for me? WAREL 05:17, 4 March 2006 (UTC)

This is easy,

${\displaystyle n\ln \ln n\leq n\ln n}$

for ${\displaystyle n\geq 6}$ so

${\displaystyle {\frac {1}{n\ln n+n\ln \ln n}}\geq {\frac {1}{2n\ln n}}}$ for ${\displaystyle n\geq 6}$.

We know that the sum

${\displaystyle \sum _{n=6}^{\infty }{\frac {1}{n\ln n}}}$

diverges by comparison with

${\displaystyle \int _{6}^{\infty }{\frac {1}{x\ln x}}\,dx=\ln \ln x|_{6}^{\infty }}$

(sorry for the informal use of the indefinite integral). —Preceding unsigned comment added by 129.128.159.46 (talk • contribs)

Here's how it can be done:

${\displaystyle \sum _{n=6}^{\infty }{\frac {1}{n\ln n+n\ln \ln n\quad }}>\sum _{n=6}^{\infty }{\frac {1}{n\ln n+n\ln n}}={\frac {1}{2}}\sum _{n=6}^{\infty }{\frac {1}{n\ln n}},}$

and then compare that with

${\displaystyle \int _{6}^{\infty }{\frac {1}{x\ln x}}\,dx.}$

This integral can be found via the substitution u = ln(x), du = dx/x, so that it becomes the integral of du/u, which clearly diverges to ∞. Michael Hardy 20:16, 25 May 2007 (UTC)

I am having no luck in following the logic of the second proof by Erdos. I don't understand how it is important that here are at most 2^i CHOICES for k. This does not mean that k < 2^i (which seems to be how this fact is being used).

I agree that the wording is poor. I think the author meant that if we want to form a number ${\displaystyle n\leq x}$ that is not divisible by any prime other than the first i, we have at most 2^i choices for the square-free part of the number (since we only have i primes to work with) and at at most ${\displaystyle {\sqrt {x}}}$ choices for the square part. Hence the reason ${\displaystyle N(x)\leq 2^{i}{\sqrt {x}}}$. —Preceding unsigned comment added by 129.128.159.46 (talk • contribs)

In the second proof by Erdos, when he shows we have at most 2^i possibilities for k, he uses this fact to put bounds on the SIZE of the set M. CecilPL (talk) 14:34, 26 June 2009 (UTC)

Why was my response to this request deleted? I believe I correctly addressed the issue. —Preceding unsigned comment added by 198.161.30.123 (talk • contribs)

Please sign talk page comments with ~~~~. You could ask User:Michael Hardy who reverted it. Maybe it was because you inserted comments inside earlier posts by other editors or IP numbers without signing. That is confusing and can be seriously misleading if people think somebody else wrote it. (In fact the reverted edits were by another IP but it sounds like it was you). It's much better to comment below others and sign. I recommend to register. One of many advantages is that people can connect your talk page posts when you sign. PrimeHunter 13:33, 28 May 2007 (UTC)

Aye, how is fifth proof correct? atul 14:15, 16 June 2007 (UTC)

A simple, elegant proof

along the Euler idea -- see pl:Liczby pierwsze, section 5.1 -- Wlod 00:45, 10 November 2007 (UTC)

The numbering may vary as editors add, remove or reorder sections. Here is a link pointing at the proper section by its name: pl:Liczby pierwsze#Szereg odwrotności wszystkich liczb pierwszych. --CiaPan 14:32, 13 November 2007 (UTC)

I think the third proof here is slightly simpler -- Schmock 16:37, 10 November 2007 (UTC)

The third proof here requires to factor the decomposition of a number into a square and a square-free part. Otherwise, I think, the third proof here has the following advantages (besides being more explicit and wikified):

1. There are no infinite series involved explicitly (in particular not the geometric series).
2. The estimate for the exponential function is easier to see, for example from the power series representation or via first and second derivatives of the exponential function.
3. The constant is closer to Meissel-Mertens constant.

-- Schmock 15:21, 13 November 2007 (UTC)

Thank you CiaPan for a clean format of the link to pl.wiki. I like that a*b^2 proof myself. Nevertheless, there is some tension between first and second argument above, while my proof involves nothing more than geometric progression. Everything is explicit, there is no hidden cost. Furthermore, my proof does not use the uniqueness of the decomposition into primes (just existence). And finally, and what is the most important, it gives, at virtually no cost, much more than just Euler's theorem, because it presents one half of the very sharp theorem by Mertens. We get the loglog estimate. This sheds light onto the topic, indicating strongly that one direction is much simpler than the other. Here is a link to my knol "Infnitude of primes - 2", section "Franz Mertens (1840-1927)", in which I presented this proof in English: http://knol.google.com/k/wlodzimierz-holsztynski/infnitude-of-primes-2/1jxfhq4x4sw0j/62#H3-Franz-Mertens-(1840-1927)

BTW, right there, in the next section, "Decomposition a * b^2 + Euler method", you'll find a combination of the a*b^2 and Euler arguments (which lead to a sharper result than more elementary a*b^2 argument alone).

(I'll address point 3, see above, later. I think that it is the other way around).

-- Wlod (talk) 19:11, 14 November 2008 (UTC)

Finally I truly read the third proof, and I like it a lot! The discussion about the relative advantages of that proof over mine is mute--mine is simpler but not necessarily "better". (I even made a cosmetic modification to the proof 3 here to make it slightly sharper--nothing major, but it is a useful moment in many similar inequalities). The Euler idea is used both to primes and to the squares (!) in proof 3, which is really very-very nice. -- Wlod (talk) 20:09, 14 November 2008 (UTC)

Let me also address the unimportant issue of the constant (Schmock's point 3 above)--unimportant for this kind of inexact, elementary proofs. With a little attention, I got the constant in my proof better than the one here, in proof three--now I have, in my knol, the simple version with constant -1, and a bit more involved version with constant better than -.381. As I said, it's not too important, I did it for my peace of mind. -- Wlod (talk) 03:38, 17 November 2008 (UTC)

An Aesthetic Question

Shouldn't the sum ${\displaystyle \sum _{p\leq n}{\frac {1}{p}}}$ appear explicitly at the start of this article? 68.150.231.179 (talk) 08:07, 10 November 2008 (UTC)

Why? ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{p_{n}}}=\infty }$ or ${\displaystyle \sum _{p}{\frac {1}{p}}=\infty }$ would seem more relevant to me. PrimeHunter (talk) 14:39, 10 November 2008 (UTC)

What I mean is that none of the above sums appear explicitly in the first few paragraphs of the article. It might not be clear to the uneducated reader what the article is actually about, especially since the first mathematical expression that is presented is the regular harmonic series. I think it might be more appropriate to show the sum near the start of the article to help illustrate the statement "sum of the reciprocals of all prime numbers diverges". 68.150.231.179 (talk) 16:46, 11 November 2008 (UTC)

Readers who don't understand "the sum of the reciprocals of all prime numbers diverges" may not understand a summation symbol either, so for that purpose this might be better:

${\displaystyle {\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{11}}+\cdots =\infty .}$

PrimeHunter (talk) 22:16, 11 November 2008 (UTC)

xkcd comic

Vis a vis the recent additions and reversions, I can't make sense of the assertion that a comic illustrating the divergence of the sum of the reciprocals of the primes isn't relevant to an article about proofs of the divergence of the reciprocals of the primes. And unless the 'in popular culture' section is turning into a laundry list, the notability criterion is typically applied to articles rather than to links between articles. (Perhaps the reverter meant that the comic itself isn't notable enough for a[REDACTED] entry?) However, if an episode of a popular math-themed comic that features a representation of the statement proven doesn't constitute a relevant and notable reference to this topic, I'm left wondering what would qualify. kraemer (talk) 18:34, 26 June 2009 (UTC)

Lots of people think nothing qualifies. The xkcd is #602; I don't happen to find it very clear. Nonetheless, I think it ought to be included, as one of the very few non-technical references this subject is likely ever to have. I've readded, as an external link; this may get less static. 21:47, 26 June 2009 (UTC)

I'm pretty sure I get that comic — the stick guy starts grabbing unit squares corresponding to the primes and then manually squishing them to perform a transformation that's invariant WRT area but fixes the height to be equal to the value of the prime. He stacks them up and observes that the height won't be a finite number. (What a nerd.) kraemer (talk) 07:27, 27 June 2009 (UTC)

Edits of Pirokiazuma

This is a blatant content degradation on the edge of WP:vandalism. Any replacement of “x” to “65” or alike shall be immediately reverted as, at best, an unsourced original research. Incnis Mrsi (talk) 07:08, 24 June 2013 (UTC)

I just kept it simple. Do you see what I wrote is correct? Pirokiazuma (talk) 07:36, 24 June 2013 (UTC)

A divergence can’t be proved on any finite number of partial sums. You have 2 hours to demonstrate something like a reliable source where a significance of M₆₅ is asserted. FYI, it does not imply that during these 2 hours your original research will not be reverted by somebody else than me. Incnis Mrsi (talk) 07:48, 24 June 2013 (UTC)

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