1. ## 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.

2. 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}}
$

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

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

3. ## 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.

4. 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.
$\text{lcm}(d_1,d_2)=\frac{d_1d_2}{\gcd(d_1,d_2)}.$

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

--Kevin C.