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I have to prove the following identity: $$\sum_{n\geq 0} (-1)^n(2n+1)q^{\frac{n(n+1)}{2}} = (q;q)_\infty^3$$ where $(a;q)_\infty = \prod_{i\geq 0}1-aq^i$ is the $q$-Pochhammer symbol. In my notes the identity is stated as a corollary of Jacobi triple product, whose statement is $$\sum_{n\in\mathbb Z} z^nq^{\frac{n(n+1)}{2}} = (-zq;q)_\infty(-1/z;q)_\infty(q;q)_\infty$$

What I have done so far is to change the range of $n$ to $\mathbb Z$, noting that

\begin{align*}\sum_{n\geq 0} (-1)^n(2n+1)q^{\frac{n(n+1)}{2}} &= \sum_{n\in\mathbb Z} (-1)^n n q^{\frac{n(n+1)}{2}} \\ &= \sum_{n\in\mathbb Z} (-1)^n \frac{2n+n^2-n^2}{2} q^{\frac{n(n+1)}{2}} \\ &= \sum_{n\in\mathbb Z} (-1)^n \frac{n^2+n}{2} q^{\frac{n(n+1)}{2}} + \sum_{n\in\mathbb Z} (-1)^n \frac{n-n^2}{2} q^{\frac{n(n+1)}{2}} \\ &= q\left(\sum_{n\in\mathbb Z} (-1)^n q^{\frac{n(n+1)}{2}}\right)' + \sum_{n\in\mathbb Z} (-1)^{n+1} \frac{n(n-1)}{2} q^{\frac{n(n+1)}{2}} \\ &= 0 + \sum_{n\in\mathbb Z} (-1)^{n+1} \frac{n(n-1)}{2} q^{\frac{n(n+1)}{2}} \end{align*} where the first sum is zero, thanks to Jacobi triple product with $z=-1$.

However, I can't tell if I'm getting closer to my goal. Any hint would be appreciated at this point!

As a side question, I think I'm not getting the Jacobi triple product properly. In the LHS, there are negative powers of $z$. Does this mean that the equality is not happening in $\mathbb C[[z,q]]$ but only for values of $z,q$ where the sum converges? If so, what is the region of convergence of the series?

ccorn
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zarathustra
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1 Answers1

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The theorem is due to Jacobi. An english-language demonstration can be found in chapter 19.9, at theorem 357 (with typos) by

  1. G. H. Hardy and E. M. Wright: An introduction to the theory of numbers. Oxford University press, sixth edition 2008, ISBN 978-0-19-921985-8. Note: First edition 1938. The fourth edition seems to be preferable, the sixth has shameful OCR artefacts.

The proof goes like this. Jacobi's triple product identity, as quoted in the question: $$\sum_{n\in\mathbb{Z}} z^n q^{n(n+1)/2} = (-zq;q)_\infty (-\tfrac{1}{z};q)_\infty (q;q)_\infty$$ To get somewhere near $(q;q)_\infty^3$, we want to set $z=-1$. The problem is then the second product: Its first factor is $(1+\tfrac{1}{z})$ which becomes zero for $z=-1$. So let us pull this first factor out: $$\begin{align} (-\tfrac{1}{z};q)_\infty &= (1+\tfrac{1}{z}) (-\tfrac{1}{z}q;q)_\infty \\\therefore\quad (-zq;q)_\infty (-\tfrac{1}{z}q;q)_\infty (q;q)_\infty &= \frac{\sum_{n\in\mathbb{Z}} z^n q^{n(n+1)/2}}{1+\tfrac{1}{z}} \\ &= \sum_{n=0}^\infty \frac{z^n+z^{-n-1}}{1+\tfrac{1}{z}} q^{n(n+1)/2} \\ &= \sum_{n=0}^\infty z^{-n}\frac{z^{2n+1}+1}{z+1} q^{n(n+1)/2} \\ &= \sum_{n=0}^\infty z^{-n}\,(z^{2n}-z^{2n-1}\pm\cdots-z+1)\, q^{n(n+1)/2} \end{align}$$ Thus, the singularity at $z=-1$ can be removed. Setting $z=-1$ now gives $$(q;q)_\infty^3 = \sum_{n=0}^\infty (-1)^n (2n+1)\,q^{n(n+1)/2}$$ which is what we wanted to prove.

ccorn
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