Riemann zeta function - one identity

  • Thread starter Karamata
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  • #1
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Let [tex]p_n[/tex] be number of Non-Isomorphic Abelian Groups by order [tex]n[/tex]. For [tex]R(s)>1[/tex] with [tex]\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}[/tex] we define Riemann zeta function. Fundamental theorem of arithmetic is biconditional with fact that [tex]\zeta(s)=\prod_{p} (1-p^{-s})^{-1}[/tex] for [tex]R(s)>1[/tex]. Proove that for [tex]R(s)>1[/tex] is: [tex]\sum_{n=1}^{\infty}\frac{p_n}{n^s}=\prod_{j=1}^{∞}\zeta(js)[/tex].

Do you know where can I found this proof (or maybe you know it :smile:)

Sorry for bad English :biggrin:
 

Answers and Replies

  • #2
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The number of abelian groups of order n = ∏piai is ∏p(ai) where p is the partition function.

The generating function for p is Ʃp(n)xn=(1+x+x2+..)(1+x2+x4+..)(1+x3+..).. and we only need to know for later how to see that (the exponent taken from the first parantheses corresponds to the number of 1s in the partition etc).

Now ∏ ζ(js) = ζ(s)ζ(2s)ζ(3s) = ∏(1+1/ps+1/p2s+..)(1+1/p2s+..).. where the last product is over all primes, and we have written the Euler product of each factor ζ(js).

Write out some more terms, then study the expression, and you will see that the coefficient of 1/ns, after multiplying, is precisely what you want.
 

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