# How to prove this equality of series

While studying some physics problems, I stumbled upon this experimental equality:

$$\sum_{k, \ell = 0}^{+\infty} q^{\frac{ 1 }{ 2 }[( k + \ell + 1)^2 - (k- \ell)]} = \frac{ \sqrt{q} }{ 1-q } \ .$$

I have checked this equality with Mathematica to a very high order: However, I wonder how to prove it? Is it (or its generalization) discussed somewhere? (the sum smells like some $$\Theta$$-function, but I don't know where to look).

Update:

It turns out that the series above equals the false theta function $$g_{1,1,1}(-1, -q, q)$$ discussed in this paper https://www.sciencedirect.com/science/article/pii/S0022314X18300611 . It remains to educate myself about these special function

• @Physor I think it's $k - \ell$, according to my Mathematica result. Oct 6, 2020 at 13:14
• @Physor But in any case, there was a typo in the original post : ) Oct 6, 2020 at 13:15
• If you find out how to sum $$\sum_j q^{j^2}$$ then you can reduce sth, but I'm not sure Oct 6, 2020 at 13:27

Rewrite $$\sum_{k, \ell = 0}^{+\infty} q^{\frac{ 1 }{ 2 }[( k + \ell + 1)^2 - (k- \ell)]} = \frac{ \sqrt{q} }{ 1-q }$$ as $$\sum_{k, \ell = 0}^{+\infty} x^{( k + \ell + 1)^2 - (k- \ell)} = \frac{x }{ 1-x^2 }$$ RHS is the sum of the series $$\sum _{n=1}^{\infty } x^{2 n-1}=\frac{x }{ 1-x^2 };\;0\le x<1$$ I say that the sequence $$\{( k + \ell + 1)^2 - (k- \ell)\};\; k,\ell\in\mathbb{N}$$ is exactly the sequence of the odd positive integers.
$$k^2+2 k \ell+k+\ell^2+3 \ell+1$$
We want to prove that, subtracting $$1$$ and dividing by $$2$$, the expression $$f(k,\ell)=\frac{1}{2}\left(k^2+2 k \ell+k+\ell^2+3 \ell\right)$$ gives, for $$k=0,1,2,\ldots;\;\ell=0,1,2,3,\ldots$$ all the natural numbers only once.
That is, the function $$f:\mathbb{N}^2\to\mathbb{N}$$ is a bijection.