How can we use the Jacobi-Triple Product Identity to show that in the area of q-analog series and theory:
Jacobi- Triple Product Identity: $$\sum_{{n\in\mathbb{Z}}}x^nq^{n^2}=\prod_{k \ge 1}(1-q^{2k})(1+xq^{2k-1})(1-x^{-1}q^{2k-1})$$
Now how to use the above to show that:
$$\prod_{k\ge1}\frac{1-q^k}{1+q^k}=\sum_{{n\in\mathbb{Z}}}(-1)^nq^{n^2}$$ $$\prod_{k\ge1}\frac{1-q^{2k}}{1-q^{2k-1}}=\sum_{n\ge0}q^{\binom{n+1}{2}}$$
I can prove the Jacobi triple product identity, but I have alittle trouble using suitable manipulations to show the following two identities from the Jacobi's identity. I would appreciate the help offered.
Put $x=-1$ and $x=q$ into these equations. Look at the second. $$\sum_{n=-\infty}^\infty q^{n^2+n}=\prod_{k=1}^\infty(1-q^{2k}) (1+q^{2k})(1+q^{2k-2}).$$ Divide by $2$ and replace $q^2$ by $q$: $$\sum_{n=0}^\infty q^{(n^2+n)/2}=\prod_{k=1}^\infty(1-q^{k})(1+q^{k})^2 =\prod_{k=1}^\infty\frac{(1-q^{2k})^2}{1-q^k}.$$ Then $\prod_{k=1}^\infty(1-q^k)=\prod_{k=1}^\infty(1-q^{2k-1}) \prod_{k=1}^\infty(1-q^{2k})$. Using this will simplify the product to your form.