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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 6 of Apostol's 'Introduction to analytic number theory'. My question is about the partial sum $L_N(\chi)$. Here the subscript $N$ is 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? I asked this question in mathematics stack exchange(here) and got a partial answer. I would like to see if there are more relevant results in this direction.

Edit: note here that the $N$ in $L_N(\chi)$ is the same as the conductor of the character $\chi$. I forgot to emphasise this point in the first version, which led to some misunderstandings.

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    $\begingroup$ Greg Martin's answer on MSE is the right one. $\endgroup$
    – reuns
    Commented Jan 16, 2023 at 13:27
  • $\begingroup$ What do you mean the answer there was a partial answer? It tells you the answer is no. What more do you want? $\endgroup$
    – Kimball
    Commented Jan 16, 2023 at 13:28
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    $\begingroup$ I believe here the parameter $m$ is taken to be the conductor of $\chi$ which is not addressed by the answer at MSE. The work of Granville-Soundararajan mentioned there says: for given $m$, one can find a quadratic character $\chi$ with $\sum_{n \le m} \chi(n)/n$ negative. In OP's problem, $m$ is the conductor of $\chi$ so the variables $m$ and $\chi$ are 'coupled'. OP's problem turns out to be much easier than the results of Granville-Soundararajan (at least conditionally). $\endgroup$ Commented Jan 16, 2023 at 13:38
  • $\begingroup$ @Kimball:in this question, I am interested in the sum of the first $N$ terms of the Dirichlet series, where $N$ is the conductor of the character, NOT an arbitrary integer independent of $\chi$. $\endgroup$
    – Zhang
    Commented Jan 16, 2023 at 13:54
  • $\begingroup$ Ah, okay, thanks for the clarification. $\endgroup$
    – Kimball
    Commented Jan 17, 2023 at 0:49

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Theorem 6.18 in the same chapter of Apostol shows, by partial summation, that $$L_m(\chi) = L(1,\chi) + O_{\chi}(m^{-1})$$ where the implied constant is effective and depends on $\chi$. This can be explicated by using the Pólya--Vinogradov inequality, which gives $$L_m(\chi) = L(1,\chi)+O( m^{-1} N^{1/2} \log N)$$ where $N$ is the conductor and the implied constant is effective and absolute (one reference is exercise 11.2.1.3 in p. 374 of Montgomery and Vaughan's book). Taking $m=N$ as you do gives $$L_N(\chi) = L(1,\chi)+O( N^{-1/2} \log N)$$ unconditionally with an absolute and effective constant. Explicit constants were worked out by various authors, see e.g. Pomerance's paper "Remarks on the Pólya-Vinogradov inequality", Integers 11, No. 4, 531-542, A19 (2011).

To conclude that this is positive, one needs a lower bound on $L(1,\chi)$. There are two lower bounds on $L(1,\chi)$: effective and non-effective.

We have $L(1,\chi)\gg N^{-1/2}$ with an effective constant, see Theorem 11.11 in the Montgomery--Vaughan book, following from a result of Page. This falls just short of being useful here, so I do not think your question can be verified unconditionally.

Siegel showed $$L(1,\chi) > C(\varepsilon)N^{-\varepsilon}$$ for any given $\varepsilon>0$. The positive constant $C(\varepsilon)$ is not effective (it is if one shows there are no Siegel zeros). See Theorem 11.14 of the same book. Taking, say, $\varepsilon=1/3$, we find $L_N(1,\chi)$ is positive for $N$ sufficiently large. Assuming no Siegel zeros, you can make this effective. So conditionally, to verify your question one needs to check finitely many values. I wouldn't be surprised if the answer is positive.

On GRH, $L(1,\chi) \gg 1/\log \log N$ (Littlewood). The constant has been made effective by Lamzouri, Li and Soundararajan in "Conditional bounds for the least quadratic non-residue and related problems" (Math. Comput. 84, No. 295, 2391-2412 (2015); corrigendum ibid. 86, No. 307, 2551-2554 (2017)), see their Theorem 1.5. This should allow you to get a quick verification of your question under GRH.

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  • $\begingroup$ What the OP fails to say is that in the original MSE question, the answer states it can be negative. $\endgroup$
    – Kimball
    Commented Jan 16, 2023 at 13:27
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    $\begingroup$ @Kimball I initially understood the question the same way Greg Martin understood it in MSE (without reading the MSE link). Upon reading what's written in the MSE link and OP's comments on Martin's answer, I understood OP's intention -- in this problem, the parameter $m$ grows with the conductor of $\chi$ (namely, it is the conductor), which does make the answer positive. (See the first version of my answer to see essentially the same answer as Martin, with some updates, since some relevant work came out very recently.) $\endgroup$ Commented Jan 16, 2023 at 13:32

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