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For $|q| < 1$, the generating function of the partition function $p(n)$ is given by $$ \sum_{n=0}^\infty p(n) q^n = \prod_{k=1}^\infty {1 \over 1-q^k}. \tag{1} $$ I have an intuitive understanding of this, but how do you know that the LHS converges?


Here's what I do know:

The RHS converges absolutely for $|q| < 1$ because its convergence is equivalent to that of $$ \sum_{k=1}^\infty \log {1 \over 1-q^k} = \sum_{k=1}^\infty O(q^k). $$ We can compare this to the partial products, since they relate to partitions in a known way, i.e. they cover all partitions where all parts are $\leq m$. So we have $$ \left| \prod_{k=1}^\infty {1 \over 1-q^k} - \prod_{k=1}^m{1 \over 1-q^k} \right| = \left| \prod_{k=m+1}^\infty p_m^*(n)q^k \right|, $$ where $p_m^*(n)$ is the number of partitions of $n$ where at least one part is $>m$. Clearly $$ \prod_{k=m+1}^\infty p_m^*(n)|q|^k \leq \prod_{k=m+1}^\infty p(n)|q|^k, $$ as long as the RHS makes sense. Letting $m \to \infty$ will complete the proof, since this last RHS tends to $0$ as long as the LHS of (1) converges.

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1 Answer 1

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I figured it out with the help of Andrews' book The Theory of Partitions. Of course in hindsight, it seems easy.

The sequence of partial sums $$ S_N = \sum_{n=0}^N p(n) |q|^n $$ is clearly increasing, and we'll show that it's also bounded. We have $$ S_N \leq \prod_{k=1}^N {1 \over 1-|q|^k} $$ because the LHS covers all partitions of all numbers $\leq N$ while the RHS covers all partitions where all the parts are $\leq N$, which clearly includes the former. Furthermore, $$ \prod_{k=1}^N {1 \over 1-|q|^k} \leq \prod_{k=1}^\infty {1 \over 1-|q|^k} $$ because all the factors are $\geq 1$. As shown in my original post, this converges, so we're done.

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You should accept your own answer! –  Bruno Joyal Mar 2 at 7:04

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