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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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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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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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