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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

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The closed form involves the Jacobi theta function... would you be alright with that? – J. M. Jul 22 '11 at 16:09
Ah sorry J.M. didn't see what you wrote - yeah wolfram alpha also gets something with theta. – Peter Sheldrick Jul 22 '11 at 16:10
In particular, see this defining series for $\vartheta_2(z,q)$ and consider the case $z=0$... – J. M. Jul 22 '11 at 16:11
up vote 3 down vote accepted

(I'm settling this cream puff before I retire for today...)


$$\vartheta_2(z,q)=2\sqrt{q}\sum_{k=0}^\infty q^\frac{k(k+1)}{2}\cos((2k+1)z)$$

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...

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