Is there any closed form known for the expression $\sum_{i=1}^\infty a^{i^2}$ where $|a|<1$? Thanks!
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6$\begingroup$ No. But look up "theta functions". $\endgroup$– Gerry MyersonCommented May 20, 2014 at 12:40
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$\begingroup$ Oh thanks. I forgot to correct the title... $\endgroup$– user51031Commented May 20, 2014 at 13:55
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2 Answers
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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.
answered May 20, 2014 at 17:02
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6$\begingroup$ Is $r_4(n)$ more "known" than $a^{n^2}$? $\endgroup$ Commented May 21, 2014 at 6:44
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$\begingroup$ @BrendanMcKay : I know what you mean, but the spirit of the question seems to be to describe the given series in terms of something "more familiar" $\endgroup$ Commented May 21, 2014 at 9:02
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$\begingroup$ 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. $\endgroup$ Commented Jan 15, 2015 at 7:24
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$\begingroup$ @HjalmarRosengren : OK, thanks, so $1+2f(a)$ is the 4-th root of a "known" function. $\endgroup$ Commented Jan 15, 2015 at 12:14
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$\begingroup$ 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. $\endgroup$ Commented Jan 15, 2015 at 12:33
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This sum equals exactly to: $$ \frac{1}{2}\left(\theta_3 (0,a) -1 \right),$$ and $\theta_3$ is the Jacobi $\theta$ - function.
