Sum of Series Where Exponent is Sum of Arithmetic Progression - MathOverflow

Sum of Series Where Exponent is Sum of Arithmetic Progression

2011-11-01T17:39:54Z

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.

Answer by John Mangual for Sum of Series Where Exponent is Sum of Arithmetic Progression

2011-11-01T18:34:26Z

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.

Answer by Robert Israel for Sum of Series Where Exponent is Sum of Arithmetic Progression

2011-11-02T00:42:02Z

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 $|x| >1$).