I know that for $|r|<1$ the infinite geometric series has an explicit value as
$$\sum_{n=0}^{\infty} r^n =\frac{1}{1-r}$$
Does there exist a similar result for
$$\sum_{n=0}^{\infty} r^{n^2}$$
I've seen some stuff on Jacobi-theta functions, but can't see how that applies to the non-complex number setting where $|r|<1$.
Will hazard a guess. Only an approximation though.
Let $\alpha=|\ln(|r|)|$.
$\sum_{n=0}^\infty \ r^{n^2}=\sum_{n=0}^\infty \ e^{-\alpha n^2}$
Now I think you can bound the sum with modifications of the integral:
$\sum_{n=0}^\infty \ e^{-\alpha n^2}\approx \int_0^\infty \ e^{-\alpha x^2} dx=\int_0^\infty \ (1-\alpha x^2+\frac{\alpha^2x^4}{2!}+...) dx$
No closed form solution for the partial sums.
