showing an identity about the dirichlet series

Question

Let $f$ be a multiplicative function. I have to show that $$\prod_{p}(\sum_{\nu=0}^\infty f(p^{\nu})p^{-\nu s})=\sum_{n=1}^\infty f(n)n^{-s}.$$

First approach : Let $n=\prod_{i=1}^r p_i^{\nu_i}$ be the unique prime factorization of $n,p_i\neq p_j ,i\neq j$. Then $$\prod_{p}(\sum_{\nu=0}^\infty f(p^{\nu})p^{-\nu s})=(\sum_{\nu_1=0}^\infty f(p_1^{\nu_1})p_1^{-s\nu_1})\cdot (\sum_{\nu_2=0}^\infty f(p_2^{\nu_2})p_2^{-s\nu_2})\cdot...\cdot (\sum_{\nu_r=0}^\infty f(p_r^{\nu_r})p_r^{-s\nu_r})=\sum_{0\le\nu_1\leq\nu_2\leq...\leq\nu_r}f(p_1^{\nu_1}p_2^{\nu_2}\cdot...\cdot p_r^{\nu_r})p_1^{-s\nu_1}p_2^{-s\nu_2}\cdot ... p_r^{-s\nu_r}=\sum_{n=1}^\infty f(n)n^{-s}.$$

Answer 1

$90$% sound ... Just an ickle problem with the highlighted plum ... \begin{eqnarray*} \prod_{p}(\sum_{\nu=0}^\infty f(p^{\nu})p^{-\nu s})=(\sum_{\nu_1=0}^\infty f(p_1^{\nu_1})p_1^{-s\nu_1})\cdot (\sum_{\nu_2=0}^\infty f(p_2^{\nu_2})p_2^{-s\nu_2}) \cdots (\sum_{\nu_r=0}^\infty f(p_r^{\nu_r})p_r^{-s\nu_r})=\sum_{\color{red}{0\le\nu_1\leq\nu_2\leq \cdots \leq\nu_r}}f(p_1^{\nu_1}p_2^{\nu_2}\cdots p_r^{\nu_r})p_1^{-s\nu_1}p_2^{-s\nu_2}\cdots p_r^{-s\nu_r}=\sum_{n=1}^\infty f(n)n^{-s}. \end{eqnarray*} There is no restriction between the $ \nu $'s. \begin{eqnarray*} \sum_{0 \le\nu_1,\nu_2 ,\cdots ,\nu_r} \end{eqnarray*} will do.

Answer 2

Try first with $f(n)=1$ it works exactly the same. For $f$ multiplicative $$\prod_{p\le k} \sum_{m\ge 0} f(p^m)p^{-sm}=\sum_{Lpf(n)\le k} f(n)n^{-s}$$ where $Lpf$ is the largest prime factor.

Thus if $\sum_n |f(n)n^{-s}|$ converges both sides are well-defined and we obtain

$$\prod_p \sum f(p^m)p^{-sm}=\lim_{k\to \infty}\sum_{Lpf(n)\le k} f(n)n^{-s}=\sum_n f(n)n^{-s}$$ If the Dirichlet series converges at some $s_0$ then it converges absolutely for $ \Re(s) > s_0$ and we have have shown its Euler product converges absolutely too, thus if the Euler factors $\sum f(p^m)p^{-sm}$ don't vanish then $\sum_n f(n)n^{-s}$ doesn't vanish.

It doesn't work away from the half-plane of absolute convergence : $\sum_n (-1)^{n+1}n^{-s}$ is entire, it converges for $\Re(s) >0$, it has plenty of zeros on $\Re(s)=1/2$, it converges absolutely for $\Re(s) > 1$, and its Euler product converges for $\Re(s) > 1$.