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Euler's Method of proving primes r infinite

  1. Nov 19, 2005 #1
    I somewhere read that Euler proved that primes are infinite by proving that the series 1/2 +1/3 + 1/5 +... diverges. Can anybody tell the proof?

  2. jcsd
  3. Nov 19, 2005 #2
    He proved it diverges, so if this series were finite, summing finite terms, how can it become infinite ? Now for the proof of the divergence, I have no idea, but you can look at : http://planetmath.org/encyclopedia/PrimeHarmonicSeries.html [Broken]
    Last edited by a moderator: May 2, 2017
  4. Nov 19, 2005 #3


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    Euler showed

    [tex]\sum_{n=1}^{\infty}\frac{1}{n^s}=\prod_{p\ prime}\left(1-\frac{1}{p^s}\right)^{-1}[/tex]

    for real values of s greater than 1. You can look at the product up to p<x say, use the forumula for a geometric series and fundamental theorem of arithmetic to expand this product. Then show this approaches the sum on the left as x->infinity.

    Let s go to 1, the sum on the left diverges (harmonic series), so the product must have infinitely many terms hence there are infinitey many primes.

    Dirichlet later showed the analagous sum for primes in arithmetic progressions diverges.
  5. Nov 21, 2005 #4
    Hey thanks Shmoe! Could you explain Dirichlet's statement(What's an analagous sum?) and proof, please?

  6. Nov 21, 2005 #5


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    Let q and a be relatively prime. The idea is to show that there are infinitely many primes of the form a+qn by showing the sum

    [tex]\sum \frac{1}{p}[/tex]

    diverges, where this sum is taken over primes p=a+qn for some value of n.

    It involves using the Dirichlet L-functions,


    where [tex]\chi(n)[/tex] is a multiplicative character from the multiplicative group [tex](\mathbb{Z}/q\mathbb{Z})^\times[/tex] to the complex numbers and extended to the naturals by periodicity and setting [tex]\chi(n)=0[/tex] if n and q are not relatively prime.

    These satisfy some nice orhtogonality relations that let us pick out arithmetic progressions like the sum we are interested in. We can show, for s>1:

    [tex]\sum\chi(a)\log L(s,\chi)=\phi(q)\sum\frac{1}{p^s}+O(1)[/tex]

    where the sum is taken over all characters mod q and the sum on the right is over all primes in our progression (phi is the usual euler phi function). When our character is trivial, [tex]L(s,\chi)[/tex] behaves much like the usual zeta function, and diverges to infinity as s approaches 1. Therefore, if you can show that the rest of the L-functions in the left hand sum behave and don't vanish (i.e. their log's behave) then the sum on the right will diverge like we need.

    The hard part turns out to be showing that [tex]\L(1,\chi)[/tex] is non zero when [tex]\chi[/tex] is a real character, that is it only takes on values in the real numbers (actually +1 or -1), but that's the basic outline.
    Last edited: Nov 21, 2005
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