What does $\sum_{n=0}^\infty z^{n(n+1)/2}$, $|z|<1$ converge to?

Question

Does anyone know what the series $$ S(z) = \sum_{n=0}^{\infty} z^\frac{n(n+1)}{2} $$ converges to for $|z|<1$? This came up in an application where $z$ is a probability and $S(z)$ an expected value. The exponent comes from the fact that $\sum_{k=0}^nk=\frac{n(n+1)}{2}$. I think $S(z)$ should definitely converge because the coefficients are nonnegative and form a subsequence of the well-known $\sum_{n=0}^\infty z^n=\frac{1}{1-z}$. Is there a closed-form representation of $S(z)$?

Answer 1

The Wikipedia article Ramanujan theta function states

$\psi(q) = f(q,q^3) = \sum_{n=0}^\infty q^{n(n+1)/2} = (q^2;q^2)_\infty(-q;q)_\infty $

and, yes, this converges only for $\,|q|<1.$ Note that his is as close to closed form as you can get. A more complicated expression for this is in terms of a Jacobi theta function.

Answer 2

Using Jacobi Elliptic theta function $$\vartheta _2(u,q)=2 \sqrt[4]{q} \sum _{n=0}^{\infty } q^{n(n+1)} \cos ((2 n+1) u)$$ we can conclude that $$\sum _{n=0}^{\infty } z^{\frac{1}{2} n (n+1)}=\frac{\vartheta _2\left(0,\sqrt{z}\right)}{2 \sqrt[8]{z}}$$ To satisfy my curiosity I plotted (for real $z$ with $|z|<1$) the graph of such an "exotic" function, below.

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