TITLE: Convergence of a family of series
QUESTION [8 upvotes]: I want to show that for $D>0, D\equiv 0,1 \pmod4$, $z \in \mathbb{C}$ with $ \operatorname{Im}(z)>0$ and $k>1$ the sum
$$\sum_{a,b,c \in \mathbb{Z} \\ b^2-4ac=D}(az^2+bz+c)^{-k}$$ converges absolutely.

REPLY [2 votes]: It is actually rather straightforward. If $a=0$, then $b$ can take at most 2 values, none of them $0$, and we are left with the absolutely convergent sum $\sum_c |c+bi|^{-k}$. If $c=0$, we can run a similar argument. So it is enough to consider $a,b,c\ne 0$. In this case the roots $u,v$ of the corresponding quadratic polynomial $P(x)=ax^2+bx+c$ are real and lie at the distance $\sqrt D$ from $-\frac b{2a}$. Thus $|P(z)|=|a||z-u||z-v|$. This is never less than $|a||\Im z|^2$ and for $|b|>(6\sqrt D+4|\Re z|)|a|$ is not less than $|a||\frac b{4a}|^2\approx \frac{|b|^2}{|a|}$. Thus, $|P(z)|\gtrsim\max(|a|,\frac{|b|^2}{|a|})\ge |b|$ and we only need to consider $\sum_{a,b,c}|b|^{-k}$. However, for fixed $b$, the number of non-zero pairs $(a,c)$ with $0\ne b^2-D=4ac$ is at most the number of divisors of $b^2-D$, which grows slower than any positive power of $|b|$.