Dirichlet series as Euler product

Question

I am trying to show the following proposition:

Let $f$ be a multiplicative function and $P$ be the set of prime numbers. Then the Dirichlet series $\sum_{n=1}^\infty f(n)/n^s$ converges to the value $\prod_{p\in P} (\sum_{m=0}^\infty f(p^m)/p^{ms}).$

My try following Serre- A Course in Arithmetic: Let $S$ be a finite set of prime numbers and let $N(S)$ be the set of positive integers all of whose prime factors belong to $S.$ Then $$\sum_{n\in N(S)} f(n)/n^s = \sum \dfrac{f(\prod_{p\in S} p^m)}{(\prod_{p\in S} p^m)^s}=\sum \dfrac{\prod_{p\in S} f(p^m)}{\prod_{p\in S} p^{ms}}=\sum \prod_{p\in S} \dfrac{f(p^m)}{p^{ms}}$$ $$=\prod_{p\in S} \sum_{m=0}^\infty \dfrac{f(p^m)}{p^{ms}}.$$ Then when $S$ increases we have the desired.

My doubts:

1- Why did he introduce the set $S$ and then he extends to infinity? Why can't we just do directly, by the Fundamental Theorem of Arithmetic? I do not see why.

2- After the first equality, the sum is running over what? What is the index of the sum? It is over $m$?

3- Each prime factor of $n\in N(S)$ has different exponents $m_i$. My impression is that Serre did the proof with equal exponents $m.$ I think I am wrong but why?

So it is easy to see that I am confused about that proof. Can anyone help me, please? (Serre says that the equality between the first term and the last term is immediate).

Answer 1

Denote $\mathrm{Re}\left(s\right)=\sigma$ for the complex variable $s\in\mathbb{C}$. I believe you must also assume that $\sum_{n=1}^{\infty}\frac{\vert f\left(n\right)\vert}{n^{\sigma}}<\infty$ for the proposition to hold. Assume this then. For a prime $p$ $$\sum_{m=0}^{\infty}\frac{\vert f\left(p^{m}\right)\vert}{p^{\sigma m}}\le\sum_{n=1}^{\infty}\frac{\vert f\left(n\right)\vert}{n^{\sigma}}<\infty$$by assumption, hence each internal sum in the RHS of your proposition is absolutely convergent. Let $z\in\mathbb{N}$ and $A_z$ be the set of $z$-smooth numbers, that is $$A_z:=\left\{ n\in\mathbb{N}:p\mid n\Rightarrow p\le z\right\} .$$Note that product of finitely many absolutely convergent can be arbitrarily rearranged, and therefore (also since $f$ is multiplicative) $$P_{z}:=\prod_{p\le z}\left(1+\frac{f\left(p\right)}{p^{s}}+\frac{f\left(p\right)}{p^{2s}}+...\right)=\sum_{n\in A_z}\frac{f\left(n\right)}{n^{s}}.$$Therefore we get $$\vert P_{z}-\sum_{n=1}^{\infty}\frac{f\left(n\right)}{n^{s}}\vert\le\sum_{n\notin A_z}\frac{\vert f\left(n\right)\vert}{n^{\sigma}}.$$Observe that if $n\le z$ then $p\mid n\Rightarrow p\le z$, so $n\in A_z$. Thus $$\sum_{n\notin A_z}\frac{\vert f\left(n\right)\vert}{n^{\sigma}}\le\sum_{n>z}\frac{\vert f\left(n\right)\vert}{n^{\sigma}}$$which tends to $0$ as $z\rightarrow\infty$, as a tail of a convergent series. Thus $P_{z}$ tends to $\sum_{n=1}^{\infty}\frac{f\left(n\right)}{n^{s}}$ as $z\rightarrow\infty$ and we're done. Edit - Further discussion regarding your question - here you have two points we need to address - the equality in terms of the arithmetic content, and also the questions regarding the convergence of the infinite product. Regarding the later question, I've addressed this above. Regarding the arithmetic content - After expanding, the generic term is $$\frac{f\left(p_{1}^{k_{1}}\right)f\left(p_{2}^{k_{2}}\right)\cdot...\cdot f\left(p_{m}^{k_{m}}\right)}{\left(p_{1}^{k_{1}}p_{2}^{k_{2}}\cdot...\cdot p_{m}^{k_{m}}\right)^{s}}$$ where $p_{i}\le z$ for all $i\le m$. Since $f$ is multiplicative, this term is of the form $\frac{f\left(n\right)}{n^{s}}$ where $n \in A_z$. Moreover, this correspondence between products of prime powers and positive integers $n$ is one-to-one, in view of the fundamental theorem of arithmetic. This explains the former.