Assume that we know $$ \prod_{k=1}^{n}(1+q^{2k-1}z)(1+q^{2k-1}z^{-1})=C_{0}+\sum_{k=1}^{n}C_{k}(z^k+z^{-k}), $$ with $$ C_{k}=q^{k^2}\frac{\prod_{j=n+k+1}^{2n}(1-q^{2j})}{\prod_{j=1}^{n-k}(1-q^{2j})}. $$ Deduce the following equation $$ \prod_{n=1}^{\infty}(1+q^{2n-1}z)(1+q^{2n-1}z^{-1})(1-q^{2n})=\sum_{n=-\infty}^{\infty}q^{n^2}z^n. $$ The problem is No.53 Pt.1 Chap.1 from Problems and Theorems in Analysis I by Pólya and Szegő.
The solution just states that we can obtain the result by taking the limit, so I tried with $$ \prod_{n=1}^{\infty}(1-q^{2n})\prod_{k=1}^{n}(1+q^{2k-1}z)(1+q^{2k-1}z^{-1})=\sum_{k=-n}^{n}q^{k^2}z^k\prod_{j=n+|k|+1}^{2n}(1-q^{2j})\prod_{j=n-|k|+1}^{\infty}(1-q^{2j}), $$ yet I don't know how to proceed from here rigorously, e.g., using the $\epsilon-\delta$ definition of limit or other properties.
I solved it recently.
Let $S_{n}$ denote the sum $$S_n=\sum_{k=-n}^{n}q^{k^2}\prod_{j=n+|k|+1}^{2n}(1-q^{2j})\prod_{j=n-|k|+1}^{\infty}(1-q^{2j})z^k.$$ Clearly we have $$ |S_n-\sum_{k=-n}^{k=n}q^{k^2}z^k|\le \sum_{k=-n}^{n}\rho^{k^2}r^k| 1-\prod_{j=n+|k|+1}^{2n}(1-q^{2j}) \prod_{j=n-|k|+1}^{\infty}(1-q^{2j}) |, $$
with $|q|=\rho<1$ and $|z|=r.$
Note that \begin{equation*} \begin{split} \delta_k &=\sum_{j=n+|k|+1}^{2n}\log(1-q^{2j})+\sum_{j=n-|k|+1}^{\infty}\log(1-q^{2j})\\ &=\sum_{j=n+|k|+1}^{2n}q^{2j}+\mathcal{O}(q^{4j})+\sum_{j=n-|k|+1}^{\infty}q^{2j}+\mathcal{O}(q^{4j})\\ &=-\sum_{j=n+|k|+1}^{2n}q^{2j}-\sum_{j=n-|k|+1}^{\infty}q^{2j}+\mathcal{O}\left(\sum_{j=n-|k|+1}^{\infty}\rho^{4j}\right)\\ &=\mathcal{O}\left(\sum_{j=n-|k|+1}^{\infty}\rho^{2j}\right) \\ & =\mathcal{O}(\rho^{2n-2|k|}),\\ \end{split} \end{equation*} so we have$$ e^{\delta_k}-1=\mathcal{O}(\rho^{2n-2|k|}), $$
\begin{equation*} \begin{split} |S_n-\sum_{k=-n}^{k=n}q^{k^2}z^k|& =\mathcal{O}\left(\rho^{2n}\sum_{k=-n}^{n}\rho^{k^2-2|k|}r^k\right) \\&=\mathcal{O}\left(\rho^{2n}\mathcal{O}(\int_{1}^{\infty}\rho^{x^2-2x}(r^x+r^{-x})dx)\right)\\ &=\mathcal{O}(\rho^{2n}), \\ \end{split} \end{equation*} since $\rho^{k^2-2|k|}r^k$ will eventually decrease monotonically when $|k|$ is large enough. Consequently we have $$ \lim_{n\to\infty}S_n=\sum_{n=-\infty}^{\infty}q^{n^2}z^n. $$
Any alternative proofs will be welcome.