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How do I get the sum of such a sequence:

$$1 + x^{-1} + x^{-3} + x^{-6} + \dotsb,$$

where the exponents are actually sum of arithmetic progression? I.e.,

$$x^{-0} + x^{-(0 + 1)} + x^{-(0 + 1 + 2)} + x^{-(0 + 1 + 2 + 3)} + \dotsb,$$

which can also be expressed as

$$\sum_{i=0}^{\infty} x^{-\frac{i(i + 1)}{2}}.$$

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    $\begingroup$ Mathematica says: $\frac{1}{2} x^{1/8} \big( \frac{2}{x^{1/8}} + \text{EllipticTheta}\big[2, 0, \frac{1}{\sqrt{x}}\big]\big)$. $\endgroup$ Nov 1, 2011 at 18:36
  • $\begingroup$ If there exists a formula (which sounds unlikely) for that one, then it could be modified to work for any arithmetic progression since it would just be a linear translation of the exponent, which would mean applying the function to $x^b$ and multiplying by $x^a$. Now, is there a closed-form formula? I doubt it. $\endgroup$
    – Will Sawin
    Nov 1, 2011 at 19:12
  • $\begingroup$ oh wait that's not true sorry. $\endgroup$
    – Will Sawin
    Nov 1, 2011 at 19:16
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    $\begingroup$ I would bet that the function is not elementary but it'll take some effort to prove it. Are you really curious about this stuff or it is just a random question? $\endgroup$
    – fedja
    Nov 1, 2011 at 23:48
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    $\begingroup$ what markov chain gave you a theta function as solution?? $\endgroup$ Nov 2, 2011 at 15:43

2 Answers 2

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I think you can get it from the Jacobi triple product identity \[ \prod_{m=1}^\infty (1-x^{2m})(1-x^{2m-1}y^2)(1+x^{2m-1}y^{-2}) = \sum_{n=-\infty}^\infty x^{n^2}y^{2n} \] We can set $x = q^{1/2}, y = q^{1/2}$. \[ \prod_{m=1}^\infty (1-q^{m})(1-q^{m+\frac{1}{2}})(1+q^{m-\frac{3}{2}}) = \sum_{n=-\infty}^\infty q^{\frac{n(n+1)}{2}} \] This is close to what you want.

Triple product identity and the theory of partition is still actively studied in many contexts.

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  • $\begingroup$ I'm going to go out on a limb here and make a guess that expressing the infinite sum as an infinite product is not what OP wanted (but is as good as OP is going to get). $\endgroup$ Nov 2, 2011 at 4:40
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The Jacobi Theta function $\text{JacobiTheta2}(z,q)$ (in Maple's notation) is $\sum _{k=0}^{\infty }2 \cos \left(\left( 2k+1 \right) z \right) { q}^{\left( 2 k+1 \right) ^{2}/4}$. So what you have is $\text{JacobiTheta2}(0,\sqrt{1/x}) x^{1/8}/2 $ (for $|x| >1$).

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