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Is there any closed form expression for the following serie?

$$\sum_{k=2}^\infty(-1)^k \left(- \frac{1}{2}\right)^{\frac{k(k+1)}{2}}$$

Or at least a proof that it is an irrational number. The context of this problem is given by the following link:

https://math.stackexchange.com/questions/2270730/whats-the-limit-of-sqrt2-sqrt2-sqrt2-sqrt2-sqrt2-sqrt2

In which it is proposed the problem of finding a closed form for the following nested radical:

$$R = \sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $$

whose signs follow the pattern: $10100100010000...$, where $1 = +$, $0 = -$. The expression for that radical is given by:

$$2\cos\left[\frac{\pi}{6}\left(1 - 2\sum_{k=2}^\infty(-1)^k \left(- \frac{1}{2}\right)^{\frac{k(k+1)}{2}}\right)\right]$$

And thus the question of obtaining a closed form expression for that series arises. As noted in the post, it resembles a Jacobi theta's function, except that the range for $k$ to sum over doesn't match.

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    $\begingroup$ It is called partial theta function. See formula $(1.6)$ in this article link.springer.com/article/10.1007/s11139-012-9370-1 , Kathrin Bringmann, Amanda Folsom, Robert C. Rhoades "Partial theta functions and mock modular forms as q-hypergeometric series". It does not have a closed form. $\endgroup$
    – Nemo
    Commented Oct 25, 2017 at 14:05
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    $\begingroup$ Incomplete theta functions are complicated, $\sum_{k=-\infty}^\infty(-1)^k k\ q^{\frac{k(k+1)}{2}}$ is a complete one. Is it possible to adapt your nested radical ? $\endgroup$
    – reuns
    Commented Oct 25, 2017 at 23:51

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