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I was trying to prove if it's true that $\sum_{n=0}^{\infty} e^{-\pi n^2} \left(1 - 4\pi n^2\right) (-1)^{\frac{1}{2} n(n+1)} = \frac{3}{2}$, an "identity" I found online.

The sum of the first $10$ terms with Mathematica is $1.499999999882174822912079...$, so obviously the identity is not true, but I would still try to find a closed form expression.

If I just consider $\sum_{n=0}^{\infty} e^{-\pi n^2} \left(1 - 4\pi n^2\right)$, is quite simple to show that, with the use of theta functions and this answer Proving $\sum_{n=-\infty}^\infty n^2e^{-\pi n^2}=\frac{\Gamma (1/4)}{4\sqrt{2}\pi^{7/4}}$, the sum reduces to

$1 + \frac{\theta_3(0,e^{-\pi})-1}{2} -4\pi \frac{\theta_3''(0,e^{-\pi})}{2} = 1 + \frac{\theta_3(0,e^{-\pi})-1}{2} -\frac{\theta_3(0,e^{-\pi})}{2}=\frac{1}{2}$

and this suffice to prove that the "identity" cannot be true.

So I expect that a closed form for the twice alternating sum can be expressed as a linear combination of $\theta_3(0,e^{-p\pi}),\theta_4(0,e^{-q\pi}) $ and their second derivatives, with $\{p,q\}$ positive integers.

How can it be done?

Mittens
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user967210
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1 Answers1

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A small observation (not an answer), but nevertheless: note that $-1=e^{i\pi}$ and hence, the required sum can be rewritten as follows: \begin{equation*} \begin{split} \sum_{n=0}^{\infty}e^{-\pi n^2}(1-4\pi n^2)(-1)^{n(n+1)/2} &=\sum_{n=0}^{\infty}(1-4\pi n^2)\exp\left\{-\pi n^2+\frac{i\pi}{2}n(n+1)\right\}= \\ &=\sum_{n=0}^{\infty}(1-4\pi n^2)e^{\pi i n^2\tau_0+2\pi inz_0}, \end{split} \end{equation*} where $\tau_0=\frac{1}{2}+i\in\{z:\operatorname{Im}\tau>0\}$ and $z_0=\frac{1}{4}$.

If one can somehow extend the summation in the formula above to $n\in\mathbb{Z}$, then one can take an advantage of the following formula for the Jacobi theta function $\vartheta(z,\tau)$: $$ \vartheta(z,\tau)=\sum_{n=-\infty}^{+\infty}e^{\pi in^2\tau+2\pi i nz}. $$ However, I am not sure at the moment how to proceed further.

richrow
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  • This likely goes under the rubric of so-called partial theta functions. (I say likely because I don't know how one accounts for the $n^2$ component.) – Semiclassical Mar 10 '25 at 22:37
  • @Semiclassical I do not know anything about the partial theta functions, but it is likely that the $n^2$ component can be obtained as a multiple of the second derviative in $z$, where $q=e^{2\pi i z}$. Not sure, if that would be a satisfactory answer for the original question, though. – richrow Mar 10 '25 at 22:41