Sum of Series Where Exponent is Sum of Arithmetic Progression - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:10:55Z http://mathoverflow.net/feeds/question/79731 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79731/sum-of-series-where-exponent-is-sum-of-arithmetic-progression Sum of Series Where Exponent is Sum of Arithmetic Progression edwin11 2011-11-01T17:39:54Z 2011-11-02T01:05:29Z <p>Hi,</p> <p>How do i get the sum of such a sequence:</p> <p>$1 + x^{-1} + x^{-3} + x^{-6} + ...$</p> <p>where the exponents are actually sum of arithmetic progression. i.e.</p> <p>$x^{-0} + x^{-(0 + 1)} + x^{-(0 + 1 + 2)} + x^{-(0 + 1 + 2 + 3)} + ...$</p> <p>which can also be expressed as</p> <p>$\sum_{i=0}^{\infty} x^{-\frac{i(i + 1)}{2}}$</p> <p>?</p> <p>Thanks.</p> http://mathoverflow.net/questions/79731/sum-of-series-where-exponent-is-sum-of-arithmetic-progression/79743#79743 Answer by John Mangual for Sum of Series Where Exponent is Sum of Arithmetic Progression John Mangual 2011-11-01T18:34:26Z 2011-11-01T18:34:26Z <p>I think you can get it from the <a href="http://en.wikipedia.org/wiki/Jacobi_triple_product" rel="nofollow">Jacobi triple product identity</a> $\prod_{m=1}^\infty (1-x^{2m})(1-x^{2m-1}y^2)(1+x^{2m-1}y^{-2}) = \sum_{n=-\infty}^\infty x^{n^2}y^{2n}$ 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. </p> <p>Triple product identity and the theory of partition is <a href="http://front.math.ucdavis.edu/search?a=&amp;t=&amp;q=triple+product+identity&amp;c=&amp;n=28" rel="nofollow">still actively studied</a> in many contexts.</p> http://mathoverflow.net/questions/79731/sum-of-series-where-exponent-is-sum-of-arithmetic-progression/79788#79788 Answer by Robert Israel for Sum of Series Where Exponent is Sum of Arithmetic Progression Robert Israel 2011-11-02T00:42:02Z 2011-11-02T01:05:29Z <p>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 $|x| >1$).</p>