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My question is about the end of the proof of theorem 1.1, in page 27.

Namely, it is stated that whenever we have a multiplicative function $f:\mathbb{N} \to \mathbb{C},$ let the sequence $\Lambda_{f}(n)$ be defined via the Dirichlet series $$-\frac{D'_{f}(s)}{D_{f}(s)}=\sum_{1}^{\infty}\Lambda_{f}(n)n^{-s}$$ where $D_{f}=\sum_{1}^{\infty} f(n) n^{-s}$ is the Dirichlet series of $f(n).$ Now let's suppose that $$\sum_{n\leq x}\Lambda_{f}(n)=k \log{x}+O(1) $$ for some constant $k>-\frac{1}{2}$ and that $$\sum_{n \leq x}|f(n)| \ll (\log x)^{|k|}$$

The fact that I am not able to justify is that under these assumptions the following infinite product $$ \prod_{p}(1-p^{-s-1})^{k}\left(\sum_{\nu=0}^{\infty}f(p^{\nu})p^{-\nu s}\right)$$ has the following limit as $s \to 0$ from the right $$\prod_{p}(1-p^{-1})^{k}\left(\sum_{\nu=0}^{\infty}f(p^{\nu})\right) $$

Any suggestions on that ?

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Notice that $\prod_{p\leq\\z}(1-p^{-1})^{k}\left(\sum_{\nu=0}^{\\r}f(p^{\nu})\right) \leq O((\log z)^{-k})\left(\sum_{n\leq\\x}f(n)\right)$, where $\\n\leq\ x$ is chosen accordingly to $\\z\geq\ p$. So, letting $\\x,z\to \infty$ we note that the product is convergent. Now, $\prod_{p}(1-p^{-s-1})^{k}\left(\sum_{\nu=0}^{\infty}f(p^{\nu})p^{-\nu s}\right) = D_f(s)/(\zeta(s+1))^k$ which is absolutely 'okay' for $\\Re(s)>0$, also $D_f(s)/(\zeta(s+1))^k$ is analytic at $s=0$ by checking limit (having no pole there) . Hence, by well known facts from complex analysis letting $\\s\to\ 0+$ we get the result.

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