I have to prove the following identity: $$\sum_{n\geq 0} (-1)^n(2n+1)q^{\frac{n(n+1)}{2}} = (q;q)_\infty^3$$ where $(a;q)_\infty = \prod_{i\geq 0}1-aq^i$ is the $q$-Pochhammer symbol. In my notes the identity is stated as a corollary of Jacobi triple product, whose statement is $$\sum_{n\in\mathbb Z} z^nq^{\frac{n(n+1)}{2}} = (-zq;q)_\infty(-1/z;q)_\infty(q;q)_\infty$$
What I have done so far is to change the range of $n$ to $\mathbb Z$, noting that
\begin{align*}\sum_{n\geq 0} (-1)^n(2n+1)q^{\frac{n(n+1)}{2}} &= \sum_{n\in\mathbb Z} (-1)^n n q^{\frac{n(n+1)}{2}} \\ &= \sum_{n\in\mathbb Z} (-1)^n \frac{2n+n^2-n^2}{2} q^{\frac{n(n+1)}{2}} \\ &= \sum_{n\in\mathbb Z} (-1)^n \frac{n^2+n}{2} q^{\frac{n(n+1)}{2}} + \sum_{n\in\mathbb Z} (-1)^n \frac{n-n^2}{2} q^{\frac{n(n+1)}{2}} \\ &= q\left(\sum_{n\in\mathbb Z} (-1)^n q^{\frac{n(n+1)}{2}}\right)' + \sum_{n\in\mathbb Z} (-1)^{n+1} \frac{n(n-1)}{2} q^{\frac{n(n+1)}{2}} \\ &= 0 + \sum_{n\in\mathbb Z} (-1)^{n+1} \frac{n(n-1)}{2} q^{\frac{n(n+1)}{2}} \end{align*} where the first sum is zero, thanks to Jacobi triple product with $z=-1$.
However, I can't tell if I'm getting closer to my goal. Any hint would be appreciated at this point!
As a side question, I think I'm not getting the Jacobi triple product properly. In the LHS, there are negative powers of $z$. Does this mean that the equality is not happening in $\mathbb C[[z,q]]$ but only for values of $z,q$ where the sum converges? If so, what is the region of convergence of the series?