Proof of Jacobi triple product by taking the limit 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.
 A: 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.
