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Math Help - Dirichlet Series Proof

  1. #1
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    Dirichlet Series Proof

    I'm not sure if this is the right forum for this question, but it seems the best fit to me. The question is as follows:
    Neglecting issues of the region of convergence, how does one prove the Dirichlet series generating function
    \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}=\sum_{n=1}^{\infty}\frac{\sigma_a(n)\sigma_b(n  )}{n^s},
    where \sigma_a(n) is the sum of the a-th powers of the divisors of n.
    I suspect that Dirichlet convolution figures in the solution, but am not sure how to apply it.

    --Kevin C.
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  2. #2
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    Quote Originally Posted by TwistedOne151 View Post
    I'm not sure if this is the right forum for this question, but it seems the best fit to me. The question is as follows:
    Neglecting issues of the region of convergence, how does one prove the Dirichlet series generating function
    \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}=\sum_{n=1}^{\infty}\frac{\sigma_a(n)\sigma_b(n  )}{n^s},
    where \sigma_a(n) is the sum of the a-th powers of the divisors of n.
    I suspect that Dirichlet convolution figures in the solution, but am not sure how to apply it.

    --Kevin C.
    let I_d(p,q) = \sum_{\gcd(j,k)=d} \frac{1}{j^p k^q}. clearly I_d(p,q) = \frac{1}{d^{p+q}}I_1(p,q). we have the follwing formula for I_d(p,q):

    (*) \ \ I_d(p,q)= \frac{\zeta(p) \zeta(q)}{d^{p+q}\zeta(p+q)}.

    Proof: \zeta(p)\zeta(q)=\sum_{\gcd(j,k) \geq 1} \frac{1}{j^p k^q}= \sum_{\ell \geq 1} \frac{1}{{\ell}^{p+q}}\sum_{\gcd(j,k)=1} \frac{1}{j^pk^q}=\zeta(p+q)I_1(p,q). \ \Box

    now solving your problem is quite easy:

    \sum_{n \geq 1} \frac{\sigma_a(n) \sigma_b(n)}{n^s}=\sum_{n \geq 1} \frac{1}{n^s} \sum_{d_1 \mid n} \frac{n^a}{d_1^a} \sum_{d_2 \mid n} \frac{n^b}{d_2^b}=\sum_{n \geq 1} \frac{1}{n^{s-a-b}} \sum_{d_1 \mid n} \frac{1}{d_1^a} \sum_{d_2 \mid n} \frac{1}{d_2^b}


    =\sum_{d_1 \geq 1} \frac{1}{d_1^a} \sum_{d_2 \geq 1} \frac{1}{d_2^b} \sum_{\text{lcm}(d_1,d_2) \mid n} \frac{1}{n^{s-a-b}}=\sum_{d_1 \geq 1} \frac{1}{d_1^a} \sum_{d_2 \geq 1} \frac{1}{d_2^b} \sum_{m \geq 1} \frac{1}{(\text{lcm}(d_1,d_2)m)^{s-a-b}}<br />

    =\zeta(s-a-b) \sum_{d_1 \geq 1} \frac{1}{d_1^{s-b}} \sum_{d_2 \geq 1} \frac{[(\gcd(d_1,d_2)]^{s-a-b}}{d_2^{s-a}}

    =\zeta(s-a-b) \sum_{d \geq 1} d^{s-a-b} \sum_{\gcd(d_1,d_2)=d} \frac{1}{d_1^{s-b}d_2^{s-a}}=\frac{\zeta(s-a-b) \zeta(s-a) \zeta(s-b)\zeta(s)}{\zeta(2s-a-b)}, by (*). \ \ \Box
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  3. #3
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    Clarification

    Thanks, NonCommAlg. Though, there is one step in your proof I can't quite follow; how did you get from
    \sum_{d_1\ge1}\frac{1}{d_1^a}\sum_{d_2\ge1}\frac{1  }{d_2^b}\sum_{m\ge1}\frac{1}{(\text{lcm}(d_1,d_2)m  )^{s-a-b}} to \zeta(s-a-b)\sum_{d_1\ge1}\frac{1}{d_1^{s-b}}\sum_{d_2\ge1}\frac{[(\gcd(d_1,d_2)]^{s-a-b}}{d_2^{s-a}}?

    --Kevin C.
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  4. #4
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    Quote Originally Posted by TwistedOne151 View Post
    Thanks, NonCommAlg. Though, there is one step in your proof I can't quite follow; how did you get from
    \sum_{d_1\ge1}\frac{1}{d_1^a}\sum_{d_2\ge1}\frac{1  }{d_2^b}\sum_{m\ge1}\frac{1}{(\text{lcm}(d_1,d_2)m  )^{s-a-b}} to \zeta(s-a-b)\sum_{d_1\ge1}\frac{1}{d_1^{s-b}}\sum_{d_2\ge1}\frac{[(\gcd(d_1,d_2)]^{s-a-b}}{d_2^{s-a}}?

    --Kevin C.
    \text{lcm}(d_1,d_2)=\frac{d_1d_2}{\gcd(d_1,d_2)}.
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  5. #5
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    Thanks again. I can't believe I missed that.

    --Kevin C.
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