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

Question

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.

Substitute $t=2^x$, consider $\sum\limits_{k=1}^\infty \frac{t^k}{e^{k^2}}$ — Norbert, Mar 02 '12 at 14:14

score 2 · Answer 1 · answered Mar 02 '12 at 14:19

2

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

answered Mar 02 '12 at 14:19

Ragib Zaman

35,127

3

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

score 2 · Accepted Answer · answered Mar 02 '12 at 15:20

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"

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

2 Answers2