# Sum of series $a^{i^2}$

Is there any closed form known for the expression $\sum_{i=1}^\infty a^{i^2}$ where $|a|<1$? Thanks!

• No. But look up "theta functions". May 20, 2014 at 12:40
• Oh thanks. I forgot to correct the title... May 20, 2014 at 13:55

Calling your function $f(a),$ it is clear that $f(a)^{4} = \sum_{n=1}^{\infty}r_{4}(n) a^{n},$ where $r_{4}(n)$ is the number of ways to express $n$ as a sum of four integer squares, as proved by Jacobi, who also gave an explicit description of $r_{4}(n)$ in terms of the divisors of $n.$ I that sense ( and really regarding $f(a)$ as a formal power series in $a,$ which is a slight abuse), $f(a)$ is the fourth root of a "known" function.
• Is $r_4(n)$ more "known" than $a^{n^2}$? May 21, 2014 at 6:44
• Actually, Jacobi's $r_4(n)$ refers to squares of integers, not positive integers, so it's $(1+2f(a))^4=\sum_{n=0}^\infty r_4(n)a^n$. That's an essentially different problem. As Sergei Points out, $1+2f(a)$ is a theta function, and Jacobi's work on 2, 4, 6 and 8 squares is based on exploiting this fact. Jan 15, 2015 at 7:24
• @HjalmarRosengren : OK, thanks, so $1+2f(a)$ is the 4-th root of a "known" function. Jan 15, 2015 at 12:14
• I you insist. But to me that is like saying $\cos(z)$ is the fourth root of the known function $\cos^4(z)$. Theta functions are quite fundamental and appear in lots of contexts apart from sums of four squares. Jan 15, 2015 at 12:33
This sum equals exactly to: $$\frac{1}{2}\left(\theta_3 (0,a) -1 \right),$$ and $\theta_3$ is the Jacobi $\theta$ - function.