$F(x) = L(1, \chi ) \log x + O(1)$

Question

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.

Answer 1

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 $$\sum_{n \le x} \frac{\sum_{d | n} \chi(d)}{n} = \sum_{md \le x}\frac{\chi(d)}{md}= \sum_{m \le x} \frac1m\sum_{d \le x/m} \frac{\chi(d)}{d} =\sum_{m \le x} \frac1m (L(1,\chi)+O(\frac{m}x))\\= L(1,\chi) \log x+O(1)$$

Answer 2

Observe that $A(n) = (U * \chi)(n)$, where $U(n) \equiv 1$, giving a product of Dirichlet series. Then apply the Euler summation formula as needed. We need to assume that $\chi \neq \chi_0$ because $L(s, \chi_0)$ has a pole at $s=1$.

I'm happy to provide more hints!

Dirichlet convolution gives

$$\sum_{n \leq x} \frac{(U * \chi)(n)}{n} = \sum_{n \leq x} \frac{1}{n} \cdot \sum_{n \leq x} \frac{\chi(n)}{n}.$$

The rightmost sum is $L(1, \chi)$ minus its tail, so we get

$$\sum_{n \leq x} \frac{1}{n} \cdot (L(1, \chi) + O(1))$$

Now you need to apply the Euler summation formula to $\sum_{n \leq x} \frac{1}{n}$.