Euler Product and Dirichlet Series

Question

I know the proof of this statement already exist, i need a link regarding the proof. The statement goes as:

Assume that the dirichlet series attached to f, i.e the series $$\sum_{n=1}^\infty \frac{f(n)}{n^s},$$ converge absolutely for all $r>r_0$. If $f$ is completely multiplicative, show that $$\sum_{n=1}^\infty \frac{f(n)}{n^s}= \prod_{p,prime}\frac{1}{1-\frac{f(p)}{p^s}}$$for all $r>r_0$.

Thank you very much!!!!

Answer 1

Let $Lpf(n)$ be the largest prime factor then $$\prod_{p \le k} \frac1{1-f(p)p^{-s}} = \prod_{p \le k} (1+\sum_{r=1}^\infty f(p^r) (p^r)^{-s}) =\sum_{n=1, Lpf(n) \le k}^\infty f(n) n^{-s}$$

For the absolute convergence do the same with $\prod_{p \le k} (1+\sum_{r=1}^\infty |f(p^r) (p^r)^{-s}|)$, that it converges as $k \to \infty$ is stronger than

$\prod_p (1+ | \frac1{1-f(p)p^{-s}}-1|)< \infty$ which is the definition of "$\prod_{p } \frac1{1-f(p)p^{-s}}$ converges absolutely"