How do I find a closed form solution of the following sum? $$\sigma=\sum_{i=1}^n \gamma^{\alpha i^2 + \beta i}=\sum_{i=1}^n \gamma^{\alpha i^2}\gamma^{\beta i}$$

My attempt:

I tried using the method which would solve $\sum\gamma^{\alpha i}$ where I would do this:

$$ \begin{aligned} \sigma =&\, \gamma^{\alpha} + \gamma^{2\alpha} + \dots + \gamma^{n\alpha} \\ \gamma^\alpha \sigma =& \hspace{2.675em} \gamma^{2\alpha} + \dots + \gamma^{n\alpha} + \gamma^{(n+1)\alpha} \end{aligned} $$

However there isn't a factor which does the same trick because of the $i^2$. Is there a different method for solving this, or is there no closed form solution?

  • 1
    $\begingroup$ By definition, $\displaystyle\sum_{n=-\infty}^\infty a^{n^2}=\theta_3(0,a)$, and $\displaystyle\sum_{n=-\infty}^\infty a^{n^2+n+\frac14}=\theta_2(0,a).~$ See $\theta$ function for more details. $\endgroup$ – Lucian Aug 29 '14 at 21:08

There is no "closed form" for this sum due to the $i^2$ term.

  • $\begingroup$ Can you prove this statement? $\endgroup$ – Alice Ryhl Aug 29 '14 at 20:40
  • $\begingroup$ OK. There is no known closed form. I don't think it's possible to prove that we will never know for sure. There may be some deep connection between $\pi$ and $e$ but we don't know. $\endgroup$ – amcalde Aug 29 '14 at 20:42
  • $\begingroup$ Mathematica and Maple fail to find one, even if $\beta$ is zero, which is enough proof for me. $\endgroup$ – amcalde Aug 29 '14 at 20:43

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