I am trying to find a simplified formula for the following products:


but it seems quite difficult to figure out. Just wondering if anyone know how to do this.

  • 1
    $\begingroup$ Generally speaking you won't get a closed form unless there is some nice cancellation. In this case consider that the coefficient of $a^k$ will be the number of ways you can write $k$ as a sum of distinct natural numbers up to and including $n$, multiplied by 2. $\endgroup$
    – Valborg
    Mar 23, 2018 at 22:44
  • $\begingroup$ These coefficients are pretty well studied, so you should be able to find a resource on how to compute the number of partitions into distinct parts pretty easily. $\endgroup$
    – Valborg
    Mar 23, 2018 at 22:46
  • $\begingroup$ Almost as a joke, at least for a simplified formula, this product is "just" $(-1;a)_{n+1}$ where appears the q-Pochhammer symbol. $\endgroup$ Mar 24, 2018 at 5:23

1 Answer 1


A closed form for the infinite product is the q-Pochhammer function : $$\prod_{n=0}^\infty(a^n+1)=(-1;a)_\infty$$ One can find some properties of this special function for example in : http://mathworld.wolfram.com/q-PochhammerSymbol.html

A more common closed form comes from the relationship with the Jacobi theta functions : $$\prod_{n=0}^\infty(a^n+1)=a^{-1/24}\frac{\vartheta_2(\frac{\pi}{6}\:,a^{1/3})}{\vartheta_2(\frac{\pi}{6}\:,a^{1/6})}$$ http://mathworld.wolfram.com/JacobiThetaFunctions.html


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .