What is $\sum r^{an^2+bn+c}$ when $|r|<1$?

Question

I was going through the properties of geometric series and the following question came to my mind. What I am willing to know is the following:

For $|r|<1$ what is the value of $\sum\limits_{n=1}^\infty r^{an^2+bn+c}$ with $a,b,c\in \mathbb R$ ?

I have searched in net but no positive response I have found so far. I tried to proceed manually but in vain. Can you please help me out on this matter ?

Thanks in advance

Answer 1

Suppose $a>0.$
Since $|r|<1$ we have $$0< r^{an^2+bn+c}\le r^{An+B}$$ for some $A\in\mathbb{R^+}$ and $\forall n\in\mathbb{N}.$
Now we can apply comparision test.

Similarly consider the case $a<0.$

Answer 2

For $a=\pm~b$ and/or $b=0$, we can use the Jacobi elliptic $\theta$ function: $$\sum_{n=-\infty}^\infty x^{n^2}=\theta_3(0,x),\qquad\sum_{n=-\infty}^\infty(-1)^n~x^{n^2}~=~\theta_4(0,x)~=~\theta_3(0,-x),$$ and $\quad\displaystyle\sum_{n=-\infty}^\infty x^{\big(n+\frac12\big)^2}~=~\sum_{n=-\infty}^\infty x^{\big(n-\frac12\big)^2}~=~\theta_2(0,x).\quad$ Alternately, we can also employ

approximations based on the value of the Gaussian integral and error function: $$\begin{align}\sum_{n=1}^\infty r^{an^2+bn+c}=-r^c+\sum_{n=0}^\infty r^{an^2+bn+c}~\approx-r^c+\int_0^\infty r^{ax^2+bx+c}~dx~=\\=-r^c~+~\dfrac12~\sqrt{-\dfrac\pi{a\ln r}}\cdot r^{-\tfrac\Delta{4a}}\cdot\text{ erfc}\bigg(\dfrac b2~\sqrt{-\dfrac{\ln r}a}~\bigg),\end{align}$$ where $\Delta=b^2-4ac.$