$F(x) = L(1, \chi ) \log x + O(1)$ I wish to prove $$F(x) = L(1, \chi ) \log x + O(1)$$
when $A(n) = \sum_{d|n} \chi (d)$ and $F(x) = \sum_{n \leq x} \frac{A(n)}{n}$
I started of course by substituting $A(n)$ in $F(x)$, which becomes a horrible double sum. 
Knowing that: $L(1, \chi) = \sum_{n=1}^{\infty} \frac{\chi(n)}{n}$
Any help appreciated.

Let $\chi(n)$ be $q$-periodic and $\sum_{n=1}^q \chi(n)=0$. For any $a \le q$ $$ \sum_{n \le x, n \equiv a \bmod q} \frac1n = \frac{\log(x)}q+C(a)+O(1/x)$$
$$\sum_{n \le x} \frac{\chi(n)}{n} = \sum_{a=1}^q \chi(a)\sum_{n \le x, n \equiv a \bmod q} \frac1n = \sum_{a=1}^q \chi(a) ( \frac{\log(x)}q+C(a)+O(1/x)) \\= \sum_{a=1}^q \chi(a) C(a)+ O(1/x)$$ 
Letting $x \to \infty$ gives
$$ \sum_{a=1}^q \chi(a) C(a) = L(1,\chi)$$
Therefore 
$$\sum_{n \le x} \frac{\sum_{d | n} \chi(d)}{n} = \sum_{md \le x}\frac{\chi(d)}{md}= \sum_{m \le x} \frac1m\sum_{d \le x/m} \frac{\chi(d)}{d} =\sum_{m \le x} \frac1m (L(1,\chi)+O(\frac{m}x))\\= L(1,\chi) \log x+O(1)$$
