Absolute convergence of Euler products and infinite series

Question

We know that given a multiplicative function $f$ for which the series $\sum_{n=1}^\infty f(n)$ converges absolutely then so does the Euler product $\prod_{p}\sum_{k=0}^\infty f(p^k)$, but does the reverse hold (at least up to conditional convergence)?

Answer 1

Suppose $f$ is multiplicative and $\prod_p \sum_k |f(p^k)|$ converges, i.e. there is $L$ such that for every $\epsilon > 0$ there exist $K,P$ such that $\left|L - \prod_{p \le P'} \sum_{k \le K'} |f(p^k)|\right| < \epsilon$ whenever $K' > K$ and $P' > P$. Now $$\prod_{p \le P_1} \sum_{k \le K_1} |f(p^k)| \le \sum_{n \le N} |f(n)| \le \prod_{p \le P_2} \sum_{k \le K_2} |f(p^k)|$$ where $P_1$ and $K_1$ are such that all positive integers $\prod_{p \le P_1} p^{k(p)}$ with all $k(p) < K_1$ are at most $N$, while $P_2$ and $K_2$ are such that all $n \le N$ are of the form $\prod_{p \le P_2} p^{k(p)}$ with all $k(p) \le K_2$. We conclude that the sum converges absolutely.