Positivity of partial Dirichlet series for a quadratic character?

Question

Let $\chi\colon(\mathbb{Z}/N\mathbb{Z})^\times\rightarrow\{\pm1\}$ be a primitive quadratic Dirichlet character of conductor $N$. For any integer $m=1,2,\cdots,\infty$, consider the partial Dirichlet series $$L_m(\chi)=L_m(1,\chi)=\sum_{n=1}^m\frac{\chi(n)}{n}.$$ It is well-known that $L_\infty(\chi)$ is positive, see chapter 7 of Apostol's 'Introduction to analytic number theory'. My question is about the partial sum $L_N(\chi)$. Here the subscript $N$ is the same as the conductor of $\chi$. I have done some verification in sagemath, and for $N\le1000$, $L_N(\chi)$ is always positive. Is it always the case for any $N$? How to prove this? Maybe this is a trivial question, but I did not manage to find a proof myself nor any reference online. Thanks for any hints!

Answer 1

This is definitely not a trivial question! but the answer is known: the partial sums can indeed become negative. You can find a full treatment of this problem by Granville and Soundararajan; here's an excerpt that's sufficient to justify this negative answer.

Let $\lambda(n)$ be the completely multiplicative function whose value on every prime is $-1$. Borwein, Ferguson, and Mossinghoff showed that the smallest positive integer such $N$ that $\sum_{n\le N} \frac{\lambda(n)}n < 0$ is $N= 72{,}185{,}376{,}951{,}205$.

Given this $N$, there exist infinitely many primes $p>N$ such that every prime $q\le N$ is a quadratic nonresidue modulo $p$. (Fix one quadratic nonresidue class modulo each such $q$, and throw in the congruence 1 (mod 4) as well; then by quadratic reciprocity, any prime $p$ in the intersection of all these residue classes has the required property.) Let $\chi$ be the quadratic character modulo one of these primes $p$. Then $\sum_{n\le N} \frac{\chi(n)}n$ is exactly the same sum, term for term, as $\sum_{n\le N} \frac{\lambda(n)}n$, and in particular is therefore negative.