# Sum of Series Where Exponent is Sum of Arithmetic Progression

Hi,

How do i get the sum of such a sequence:

$1 + x^{-1} + x^{-3} + x^{-6} + ...$

where the exponents are actually sum of arithmetic progression. i.e.

$x^{-0} + x^{-(0 + 1)} + x^{-(0 + 1 + 2)} + x^{-(0 + 1 + 2 + 3)} + ...$

which can also be expressed as

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

?

Thanks.

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Mathematica says: $\frac{1}{2} x^{1/8} \big( \frac{2}{x^{1/8}} + \text{EllipticTheta}\big[2, 0, \frac{1}{\sqrt{x}}\big]\big)$. –  Micah Milinovich Nov 1 '11 at 18:36
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. –  Will Sawin Nov 1 '11 at 19:12
oh wait that's not true sorry. –  Will Sawin Nov 1 '11 at 19:16
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? –  fedja Nov 1 '11 at 23:48
what markov chain gave you a theta function as solution?? –  john mangual Nov 2 '11 at 15:43

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.
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 $|x| >1$).