Compute $\frac{L'(\chi,s)}{L(\chi,s)}$.

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Let $\chi$ a character of Dirichlet mod $q$ and $$L(\chi,s)=\sum_{n\geq 1}\frac{\chi(n)}{n^s}$$ where $\Re(s)>1$. I recall that $\frac{f'}{f}:=\log(f)'$. I want to compute $-\frac{L'}{L}$. So,

$$-\frac{L'(\chi,s)}{L(\chi,s)}=\left(\log\left(\sum_{n\geq 1}\frac{\chi(n)}{n^s}\right)\right)'=\frac{\sum_{n\geq 1}\frac{\log(p)\chi(n)}{n^s}}{\sum_{n\geq 1}\frac{\chi(n)}{n^s}},$$

but how can I get something like $$\sum_{n\geq 1}\frac{f(n)}{n^s}\ \ ?$$ Because I'm suppose to arrive to something like that.

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You need to use the Euler product: $$L(\chi,s)=\prod_p\left(1-\frac{\chi(p)}{p^s}\right)^{-1}.$$ Using logarithmic differentiation gives $$\frac{L'(\chi,s)}{L(\chi,s)}=\sum_p(\ln p)\frac{\chi(p)/p^s}{1-\chi(p)/p^s}=\sum_p\sum_{n=1}^\infty\frac{\chi(p)^n\ln p}{p^{ns}}. $$