Jacobi Triple product cases Hello I want some guidance how can I prove these identities of Jacobi Tripple Product
$$\sum_{n=-\infty}^\infty q^{2n^2+n}=\prod_{n=1}^\infty\frac{(1-q^{2n})^2}{(1-q^n)}$$
$$\sum_{n=-\infty}^\infty q^{n^2}=\prod_{n=1}^\infty\frac{(1-(-q)^n)}{(1+(-q)^n)}$$ 
$$\sum_{n=-\infty}^\infty q^{n(n+1)}=\prod_{n=1}^\infty\frac{(1-q^{4n})}{(1-q^{4n-2})}$$ 
 A: Start with the triple product identity

$$\sum_{n = -\infty}^\infty z^n q^{n^2} = \prod_{n = 1}^\infty (1 - q^{2n})(1 + zq^{2n-1})(1 + z^{-1}q^{2n-1}).$$



*

*Replacing $q$ by $q^2$, then setting $z = q$, we obtain


\begin{align}
\sum_{n = -\infty}^\infty q^{2n^2+n} &= \prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n-1})(1 + q^{4n-3})\\
& = \prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{2n-1})\\
& = \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n})(1 + q^{2n-1})\\
& = \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^n)\\
& = \prod_{n = 1}^\infty \frac{(1 - q^{2n})^2}{1 - q^n}.
\end{align}


*Setting $z = 1$, we get


\begin{align}\sum_{n = -\infty}^\infty q^{n^2} &= \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n-1})(1 + q^{2n-1})\\
&= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n\, \text{odd}}^\infty (1 + q^n)\\
&= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 + q^n}{1 + q^{2n}}\\
&= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 - q^{2n}}{(1 - q^n)(1 + q^{2n})}\\
&= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 - q^{2n}}{(1 - q^{2n})(1 - q^{2n-1})(1 + q^{2n})}\\
&= \prod_{n = 1}^\infty \frac{1 - (-q)^n}{(1 - q^{2n-1})(1 + q^{2n})}\\
&= \prod_{n = 1}^\infty \frac{1 - (-q)^n}{1 + (-q)^n}.
\end{align}


*Setting $z = q$ results in


\begin{align}\sum_{n = -\infty}^\infty q^{n^2 + n} &= \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n})(1 + q^{2n-2})\\
& = \prod_{n = 1}^\infty (1 - q^{4n})\prod_{n = 2}^\infty (1 + q^{2n-2})\\
& = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty (1 + q^{2n})\\
& = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty \frac{1 - q^{4n}}{1 - q^{2n}}\\
& = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty \frac{1 - q^{4n}}{(1 - q^{4n})(1 - q^{4n-2})}\\
&= \prod_{n = 1}^\infty \frac{1 - q^{4n}}{1 - q^{4n-2}}.
\end{align}
