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Showing $prod_{ngeq 1} (1+q^{2n}) = 1 + sum_{ngeq 1} frac{q^{n(n+1)}}{prod_{i=1}^n (1-q^{2i})}$

I want to show

begin{align}
prod_{ngeq 1} (1+q^{2n}) = 1 + sum_{ngeq 1} frac{q^{n(n+1)}}{prod_{i=1}^n (1-q^{2i})}
end{align}

I know one proof via self-conjugation of partition functions with Young Tableaux. But it seems not natural for me. [In the process of the proof it appears Durfee square, etc]

Is there any other (simple?) proof for this equality?

Mathematics Asked by phy_math on December 29, 2020

1 Answers

One Answer

Substituting $sqrt{q}$ for $q$, we get $$ prod_{n=1}^{infty}(1+q^n)=sum_{k=0}^{infty}{frac{q^{binom{k+1}{2}}}{prod_{i=1}^{k}{(1-q^i)}}}. $$

The left-hand side is the generating function for partitions with distinct parts (a part of each size $n$ occurs $0$ or $1$ times).

On the other hand, if a partition with distinct parts has $k$ parts, then the $i$th smallest part is of size at least $i$. Given a partition $0<lambda_1<lambda_2<dots<lambda_k$, subtract $i$ from the size of the $i$th smallest part to get a partition with parts $0lelambda_1-1lelambda_2-2ledotslelambda_k-k$ with $le k$ parts (after exluding the $0$s), whose conjugate is a partition with the largest part $le k$. The "staircase" we subtracted has $1+2+dots+k=binom{k+1}{2}$ cells. That yields the summand on the right-hand side: $$ q^{binom{k+1}{2}}prod_{i=1}^{k}{frac{1}{1-q^i}}. $$

I think this is about as simple as it gets. The Durfee square need not be involved, as you can see.

Correct answer by Alexander Burstein on December 29, 2020

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