Sum of an infinite geometric series with squared powers

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 $$|r|<1$$.

Let $$\alpha=|\ln(|r|)|$$.
$$\sum_{n=0}^\infty \ r^{n^2}=\sum_{n=0}^\infty \ e^{-\alpha n^2}$$
$$\sum_{n=0}^\infty \ e^{-\alpha n^2}\approx \int_0^\infty \ e^{-\alpha x^2} dx=\int_0^\infty \ (1-\alpha x^2+\frac{\alpha^2x^4}{2!}+...) dx$$