Before I can ask my question, I have to state a couple of definitions. Let $f$ be a multiplicative function and let $$ D_f(s) = \sum_1^{\infty} \frac{f(n)}{n^s}, $$ and define $\Lambda_f(n)$ as follows: $$ - \frac{D'_f(s)}{D_f(s)} = \sum_1^{\infty} \frac{\Lambda_f(n)}{n^s}. $$ Suppose we have \begin{equation} \sum_{n \leq x} \Lambda_f(n) = c \log x + O(1) \end{equation} for some positive constant $c.$
I was reading a textbook and it says that the following Euler product $$ \prod_p (1 - p^{-s-1})^c (\sum_{v=0}^{\infty} f(p^v) p^{-vs} ) $$ has limit as $s \rightarrow 0$ by virtue of the above equation (third equation in this question).
I am not quite seeing how the above equation is helpful in getting that the Euler product has a limit as $s \rightarrow 0$. I would appreciate any explanation! Thank you!