I have a question dealing with the following ANT assignment:
If $f$ is multiplicative with $|f(n)|\le1$ and $F(s)=\sum_{n\ge1}\frac{f(n)}{n^s}$ is analytic around $s=1$, show that $F(s)$ has a pole at $s=1$ iff $\mathbb{D}(f,1,\infty)<\infty$ where $\mathbb{D}(f,g,x)^{2}=\sum_{p\le x}\frac{1-\Re f(p)\overline{g}(p)}{p}$
I got really stuck on this one, the main problem I have is regarding the issues of moving between summing over primes and positive integers, since convergence on one set does not guarantee anything regarding convergence on the other (meaning $\sum_p\frac{f(p)}{p}$ says nothing about $F(1)$ at least as far as I can tell). for $\Leftarrow$ assuming $\mathbb{D}(f,1,\infty)<\infty$ we find that $\sum_{p}\frac{1-\Re{f(p)}}{p}<\infty\Rightarrow \sum_{p}\frac{1-|f(p)|}{p}<\infty\Rightarrow \sum_{p}\frac{|f(p)|}{p}=\infty\Rightarrow \sum_{n}\frac{|f(n)|}{n}=\infty$ but again this is insufficient to help solve my problem.
Bottom line : I'd appreciate it if anyone could give me a better direction for solving the problem (or maybe this could work but I'm not seeing something here) Thanks