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Let $D$ be an integer number and let $\chi$ be the Dirichlet character defined by $$\chi(m) = 0 \text{ if $m$ even, } \chi(m) = (D/m) \text{ if $m$ odd,}$$ where $(D/m)$ denotes the Jacobi symbol. Let us denote by $\zeta_D(s)$ its L-series: $$ \zeta_D(s) = L(s,\chi) = \sum_{n = 1}^\infty \frac{\chi(n)}{n^s} = \prod_p \frac{1}{1-(D/p)p^{-s}},$$ where $(D/p) = 0$ if $p | 2d$.

For $s > 1$ there are the following bounds: $$\frac{\zeta(2s)}{\zeta(s)} = \prod_p \frac{1}{1+p^{-s}} \leq \zeta_D(s) \leq \prod_p \frac{1}{1-p^{-s}} = \zeta(s).$$ Both of them become trivial for $s = 1$. Still, $\zeta_D(1)$ is a finite positive number for $D < 0$.

Consider the sequence $$ \zeta_{-d}(1),\, d = 1,2,3,\ldots$$ Is this sequence bounded above or below? If yes, what is an upper or lower bound?

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    $\begingroup$ Dirichlet $L$-functions at $1$ can get arbitrarily large or small. Search under "extreme values of Dirichlet L-functions". $\endgroup$
    – Lucia
    Commented Mar 23, 2019 at 20:09
  • $\begingroup$ For $D$ odd $\chi(m) = (D/m)$ is a primitive character modulo $D$ or $4D$ why do you replace it by the non primitive character $ (D/m) 1_{gcd(m,2)=1}$. And the Siegel zero problem is exactly about that $\endgroup$
    – reuns
    Commented Mar 24, 2019 at 14:10
  • $\begingroup$ Unless there is a good reason to assume the opposite, you probably should not expect this sequence to be bounded. Generally speaking, there is a rather large proportion of Dirichlet L-function (with a given conductor) which has a "large" value at 1. See for example Theorem 2 in this recent paper: arxiv.org/abs/1803.00760 $\endgroup$
    – Kurisuto Asutora
    Commented Mar 28, 2019 at 8:27

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