# Infinite Sum of one half to triangular powers

I'm trying to figure out what the infinite sum of $\dfrac 1{2^{n(n+1)/2}}$ is.

That's $\dfrac 12 + \dfrac 1{2^3} + \dfrac 1{2^6} + \dfrac 1{2^{10}} + \cdots$ where the powers $1,3,6,10,\cdots$ are the triangular numbers.

It's convergent as its bounded above by the sum of $\left(\dfrac 12\right)^n = 1$.

I've noticed that the ration of the terms, that is $\dfrac{a_{n+1}}{a_n} = \left(\dfrac 12\right)^n$, and so the ratios form a geometric series, but can't seem to find a way to possibly use this.

I was wondering if anyone could give me a hint or a push in the right direction? Thanks

$$\vartheta_2(z,q)=2\sqrt{q}\sum_{k=0}^\infty q^\frac{k(k+1)}{2}\cos((2k+1)z)$$
be the second Jacobi theta function. Letting $z=0$ and $q=\frac12$ gives
$$\vartheta_2\left(0,\frac12\right)=\sqrt{2}\left(1+\sum_{k=1}^\infty \left(\frac12\right)^\frac{k(k+1)}{2}\right)$$
and thus the expression you want is $\frac1{\sqrt{2}}\vartheta_2\left(0,\frac12\right)-1$. I know not of any more elementary functions that can represent your sum...