How to prove the $q$-series identity?

Question

How to prove the following identity:

$$\sum_{n\ge0}\frac{2q^{n^{2}+n}}{(q)_{n}^{2}(1+q^{n})}=\sum_{n\ge0}\frac{q^{n^{2}+n}}{(q)_{n}^{2}(1-q^{2n+2})}$$

Answer 1

First, \begin{align} \sum_{0\leq n}\frac{q^{n^2+n}}{(q)_n^2(1-q^{2n+2})}&=\sum_{1\leq n}\frac{q^{n^2-n}}{(q)_{n-1}^2(1-q^{2n})}\\ &=\sum_{0\leq n}\frac{q^{n^2-n}(1-q^n)^2}{(q)_n^2(1-q^{2n})}\\ &=\sum_{0\leq n}\frac{q^{n^2-n}(1-q^n)}{(q)_n^2(1+q^n)}\\ &=\sum_{0\leq n}\frac{q^{n^2-n}}{(q)_n^2}-2\sum_{0\leq n}\frac{q^{n^2}}{(q)_n^2(1+q^n)}\\ \end{align} And $$ 2\sum_{0\leq n}\frac{q^{n^2+n}}{(q)_n^2(1+q^n)}+2\sum_{0\leq n}\frac{q^{n^2}}{(q)_n^2(1+q^n)}=2\sum_{0\leq n}\frac{q^{n^2}}{(q)_n^2} $$ So it is sufficient to show the following. $$ 2\sum_{0\leq n}\frac{q^{n^2}}{(q)_n^2}=\sum_{0\leq n}\frac{q^{n^2-n}}{(q)_n^2} $$ Use Heine's summation formula, $$ \sum_{0\leq n}\frac{(a)_n(b)_n}{(c)_n(q)_n}\left(\frac{c}{ab}\right)^n=\frac{(c/a)_{\infty}(c/b)_{\infty}}{(c)_{\infty}(c/ab)_{\infty}} $$ Take the limit of $a, b\to\infty$, we get the following. $$ \sum_{0\leq n}\frac{q^{n^2-n}}{(c)_n(q)_n}c^n=\frac 1{(c)_{\infty}} $$ Substitue $c=q$ $$ \sum_{0\leq n}\frac{q^{n^2}}{(q)_n^2}=\frac 1{(q)_{\infty}}\tag{1} $$ Also, multiply $1-c$ and $c\to 1$ \begin{align} \lim_{c\to 1}\sum_{0\leq n}\frac{q^{n^2-n}}{(cq)_{n-1}(q)_n}c^n&=\lim_{c\to 1}\frac 1{(cq)_{\infty}}\\ \sum_{0\leq n}\frac{q^{n^2-n}(1-q^n)}{(q)_n^2}&=\frac 1{(q)_{\infty}}\\ \end{align} Hence, \begin{align} \sum_{0\leq n}\frac{q^{n^2-n}}{(q)_n^2}&=\frac 1{(q)_{\infty}}+\sum_{0\leq n}\frac{q^{n^2}}{(q)_n^2}\\ &=2\sum_{0\leq n}\frac{q^{n^2}}{(q)_n^2} \end{align}