 Solution

One form of the Jacobi triple product is, for $0<|q|<1$ and $z\ne 0$,Both sides converge locally uniformly on the punctured $z$ plane. Here is a finite-product proof which also determines the multiplicative constant.

Put $Q=q^2$ and $(Q;Q{)}_m={\prod }_{j=1}^m(1-Q^j)$, with value one at $m=0$. Define the Gaussian binomial coefficient ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q=(Q;Q{)}_m/[(Q;Q{)}_k(Q;Q{)}_{m-k}]$. The recurrence ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q={\left[\genfrac{}{}{0ex}{}{m-1}{k}\right]}_Q+Q^{m-k}{\left[\genfrac{}{}{0ex}{}{m-1}{k-1}\right]}_Q$ proves by induction thatRewrite the two finite factors asThe finite identity, with $n=k-N$, makes this ${\sum }_{n=-N}^N{z}^n{q}^{n^2}{\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q$. Multiply by $(Q;Q{)}_N$. For each fixed integer $n$, the resulting coefficient tends to one, because $(Q;Q{)}_m$ tends to a nonzero limit and ${\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q\to 1/(Q;Q{)}_{\infty }$.

All finite $(Q;Q{)}_m$ are uniformly bounded above and away from zero. Thus these coefficients have one uniform bound, and ${\sum }_{n\in \mathbb{Z}}|q{|}^{n^2}|z{|}^n$ is uniformly summable on compact annuli. Dominated convergence passes the limit through the finite sums and proves the boxed formula. At $q=0$ the formula follows by its limiting value one. The frequently used alternative form isIt follows by the substitution $q^2=t$, $z=u/q$, and is the version used for the charge-counting identity in Question 8.