Dirichlet Series Proof

**TwistedOne151** · August 28th 2009, 08:01 PM

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.

**NonCommAlg** · August 29th 2009, 06:04 PM

Originally Posted by TwistedOne151

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 $(*). \ \ \Box$

**TwistedOne151** · August 29th 2009, 08:44 PM

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.

**NonCommAlg** · August 29th 2009, 09:29 PM

Originally Posted by TwistedOne151

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.

$\text{lcm}(d_1,d_2)=\frac{d_1d_2}{\gcd(d_1,d_2)}.$

**TwistedOne151** · August 30th 2009, 12:04 AM

Thanks again. I can't believe I missed that.

--Kevin C.

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