# Proof that the series for the generating function of the partition function converges?

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.

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.