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Gauss came up with some bizarre identities, namely $ \sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}=\prod_{k\geq 1}\frac{1-q^k}{1+q^k}. $

How can this be interpreted combinatorially? It strikes me as being similar to many partition identities. Thanks.

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    @GrigoryM Do you know where I can see it derived from the Jacobi triple product?2016-02-21

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There is a classical proof by Andrews which you can find in my survey here (section 5.5). There is also a bijective proof of a more general identity I gave in this paper (section 2.2). Enjoy!

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    Link fixed.....2017-08-02
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Here's a combinatorial interpretation, but I have no idea how to turn it into a combinatorial proof.

$\prod(1+q^k)$ is the generating function that counts the number of partitions into distinct parts. $\prod(1-q^k)$ is the generating function that counts the excess of partitions into an even number of distinct parts over the partitions into an odd number of distinct parts. Thus

$\sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}\prod_{k\geq 1}\left(1+q^k\right)=\prod_{k\geq 1}\left(1-q^k\right)$

considers partitions into a square and distinct parts and states that the excess of such partitions with an odd square over such partitions with an even square (where each non-zero square occurs in two colours, positive and negative) is equal to the excess of partitions into an even number of distinct parts over the partitions into an odd number of distinct parts.

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    Thanks jor$i$ki! My gut feeling is that a combinatorial proof might be quite difficult.2012-02-20
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The typical analytic proof is not difficult and is an easy consequence of Jacobi's triple product $\sum_{n=-\infty} ^{\infty} z^{n} q^{n^{2}}=\prod_{n=1}^{\infty}(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})$ for all $z, q$ with $z\neq 0,|q|<1$. Let's put $z=-1$ to get the sum in question. The corresponding product is equal to $\prod(1-q^{2n})(1-q^{2n-1})^{2}=\prod(1-q^{n})(1-q^{2n-1})=\prod \frac{(1-q^{2n})(1-q^{2n-1})} {1+q^{n}}=\prod\frac{1-q^{n}} {1+q^{n}}$ which completes the proof.

The proof for Jacobi's triple product is non-trivial / non-obvious and you may have a look at the original proof by Jacobi in this blog post.


On the other hand Franklin obtained a nice and easy combinatorial proof of the Euler's Pentagonal theorem which is equivalent to Jacobi Triple Product.

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    @reuns: Jacobi's proof in second part is a gem and I try to advertise it everywhere on MSE. This is how proofs should be. Simple and clear and requiring least mathematical machinery. Mathematical enjoyment should not be restricted to elite few University dons and instead it should cater to masses like music/art2017-08-23