Closed formula for the sum $a^1+a^4+a^9...$

Question

I'm wondering if there is a closed formula for the sum $a^1+a^4+a^9...$ and more generally $a^{1^n}+a^{2^n}+a^{3^n}...$ for real $a$ and $n$ such that $|a|<1$ and $n>1$.

Answer 1

According to Jacobi Triple product formula we have

$$\prod_{m=1}^{\infty}(1-x^{2m})(1+x^{2m-1}y^2)\left(1+\frac{x^{2m-1}}{y^2}\right)=\sum_{n=-\infty}^{\infty}x^{n^2}y^{2n}$$ for $|x|<1$ and $y\neq0$. Then taking $x=a$ and $y=1$ we get $$\prod_{m=1}^{\infty}(1-a^{2m})(1+a^{2m-1}1^2)\left(1+\frac{x^{2m-1}}{1^2}\right)=\sum_{n=-\infty}^{\infty}a^{n^2}1^{2n}$$

$$\implies\prod_{m=1}^{\infty}(1-a^{2m})(1+a^{2m-1})^2=2\sum_{n=1}^{\infty}a^{n^2}+1$$

$$\implies\sum_{n=1}^{\infty}a^{n^2}=\frac{\prod_{m=1}^{\infty}(1-a^{2m})(1+a^{2m-1})^2-1}{2}$$

You can find more things here http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.68.6437&rep=rep1&type=pdf