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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.

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

2 Answers 2

up vote 2 down vote accepted

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"

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$$\sum_{k=1}^{\infty}\frac{2^{kx}}{e^{k^{2}}} = -\frac{1}{2} + \frac{1}{2} \prod_{m=1}^{\infty} \left( 1 - \frac{1}{e^{2m}} \right) \left( 1+ \frac{ 2^x }{e^{2m-1} } \right) \left( 1 + \frac{1}{2^x e^{2m-1} }\right ). $$

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out of the frying pan into the fire –  userNaN Mar 2 '12 at 14:23
@Norbert haha yea pretty much =] But the OP wanted something, so there it is. –  Ragib Zaman Mar 2 '12 at 14:26
@Ragib: By what method did you find this? Just a name or reference please. –  Antonio Vargas Mar 2 '12 at 14:52
@AntonioVargas Google "Jacobi Triple Product". –  Ragib Zaman Mar 2 '12 at 14:53
@Ragib: Thanks! –  Antonio Vargas Mar 2 '12 at 15:02

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