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I am having a trouble understanding some equality in the book "The Theory of the Riemann zeta-function".

On pp. 9, I read \[ \sum_{n \geq 1} \frac{\sigma_{a}(kn)}{n^{s}} = \prod_{p} \sum_{m \geq 0} \frac{\sigma_{a}(p^{l + m})}{p^{ms}}, \] where $k = \prod_{p}p^{l}$. But simply expanding on the right, it seems there are many terms that I can't find in the left series.

In addition, if the relation next to it, namely, \[ \sum_{n \geq 1} \frac{\sigma_{a}(kn)}{n^{s}} = \zeta(s) \zeta(s - a) \prod_{p|k} (\sum_{m \geq 0} \sigma_{a}(p^{l + m})/p^{ms}) / \sum_{m \geq 0}\sigma_{a}(p^{m})/p^{ms}) \] were correct, then I would have \[ \sum_{n \geq 1} \frac{\sigma_{a}(kn)}{n^{s}} = \prod_{p \quad not| k} ( \sum_{m \geq 0}\sigma_{a}(p^{m})/p^{ms}) \prod_{p|k} (\sum_{m \geq 0} \sigma_{a}(p^{l + m})/p^{ms}), \] but this seems to be hard to verify too.

An explanation on this part would be really appreciated.

Thanks.

Grown pains
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  • This is a standard fact for Dirichlet series: If $f \colon \mathbb{N} \rightarrow \mathbb{C}$ is a multiplicative function, i.e. $f(ab) = f(a) f(b)$ if $\mathrm{gcd}(a,b)=1$, then (under the assumpation of absolute convergence) we have $$\sum_{n=1}^\infty\frac{f(n)}{n^s} = \prod_{p \in \mathbb{P}} \sum_{m=0}^\infty \frac{f(p^m)}{p^{ms}}$$. In order to prove this, one uses the existence and uniqueness of prime factorization and absolute convergence to reorder the sum. The same argument applies in your situation. – p4sch Mar 04 '18 at 09:28
  • Should I see the function $f(n) = \sigma_{a}(kn)$ as a multiplicative function? For an arithmetical function to be multiplicative, we must have f(1) = 1. But the f doesn't satisfy this. – Grown pains Mar 04 '18 at 13:11
  • You should use the same argument as in the proof of the above-mentioned formula (for a complete proof see e.g. Theorem 11.6 in this summary): Decompose $n = \prod_{p_i} p_i^{e_i}$ into his prime factors with $e_i =0$ for almost all $i \in \mathbb{N}$. Then $f(n) = \prod_{p_i} \sigma_a(p_i^{l_i+e_i})$. Now, use the existence and uniqueness of prime factorization to rewrite the sum as a product over the primes. – p4sch Mar 05 '18 at 09:24
  • The $k$ is some fixed number, isn't it? But then how come $f(n) = \sigma_{a}(kn) = \prod_{p_{i}|n} \sigma_{a}(p_{i}^{l_{i} + e_{i}})$ even for primes which may not divide $k$? – Grown pains Mar 09 '18 at 23:49
  • Yes, $k$ is fixed. Just decompose $k$ and $n$ into their prime-factorizations and use the multiplicity of $\sigma_a$. If a prime don't divide $k$, then it is coprime to $k$. Thus, we can use the multiplicity of $\sigma_a$. – p4sch Mar 10 '18 at 15:01
  • I just don't get it. Think it this way. Going back to $\sum_{n \geq 1} \frac{\sigma_{a}(kn)}{n^{s}} = \prod_{p}\sum_{m \geq 0} \frac{\sigma_{a}(p^{l + m})}{p^{ms}}$, suppose it is correct. Then as you say, let's use the Euler product thing, and take a closer look at how a particular $m$-term is made up on the right sum. It looks like $\sigma_{a}(p_{n_{1}}^{l_{1} + e_{1}}) \cdots \sigma_{a}(p_{n_{1}}^{l_{r} + e_{r}}) /(p_{n_{1}}^{e^{1}} \cdots p_{n_{r}}^{e_{r}})^s = \sigma_{a}[m ( p_{n_{1}}^{l_{1} } \cdots p_{n_{1}}^{l_{r}})] /m^{s} $. Then compare this with the $m$-term on the left side. – Grown pains Mar 11 '18 at 23:04
  • (continuation) How could that possibly be equal to $\sigma_{a}(km)/m^{s}$ for a ${\it fixed}$ $k$? – Grown pains Mar 11 '18 at 23:07

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