# What's the sum of $\sum_{k=1}^\infty \frac{2^{kx}}{e^{k^2}}$?

I already asked a similar question on another post: What's the sum of $\sum \limits_{k=1}^{\infty}\frac{t^{k}}{k^{k}}$? There are no problems with establishing a convergence for this power series: $$\sum_{k=1}^\infty \frac{2^{kx}}{e^{k^2}}$$ but I have problems in determining its sum.

-
I don't think this will be any less intractable... – Alex Becker Mar 2 '12 at 13:54
Substitute $t=2^x$, consider $\sum\limits_{k=1}^\infty \frac{t^k}{e^{k^2}}$ – Norbert Mar 2 '12 at 14:14

There is this Jacobi theta function: $$\vartheta_3 \biggl(\frac{i}{2} x \operatorname{ln} (2),\operatorname{e} ^{-1}\biggr) = \sum_{k = -\infty}^{\infty} \operatorname{e} ^{-k^{2}} 2^{k x}$$ But you stopped half-way through, so yours is not such a common one. Yours is a "partial theta function"
$$\sum_{k=1}^{\infty}\frac{2^{kx}}{e^{k^{2}}} = -\frac{1}{2} + \frac{1}{2} \prod_{m=1}^{\infty} \left( 1 - \frac{1}{e^{2m}} \right) \left( 1+ \frac{ 2^x }{e^{2m-1} } \right) \left( 1 + \frac{1}{2^x e^{2m-1} }\right ).$$