Recall that a $q$-Pochhammer symbol is defined as $$ (x)_n = (x;q)_n := \prod_{l=0}^{n-1}(1-q^l x). $$

I found the following curious $q$-series identity that seems to hold for any $n\geq 0$: $$ -(-1)^{n}q^{\frac{n(3n+5)}{2}}\sum_{j\geq 0}q^{j+1}(q^{j+1})_{n}(q^{j+2n+2})_{\infty} \overset{?}{=} \sum_{\substack{k\in \mathbb{Z}\\ |k| > n}}(-1)^{k}q^{\frac{k(3k+1)}{2}}. $$ Note, the right-hand side is a truncated version of the Euler function $$ \phi(q) := (q)_{\infty} = \sum_{k\in \mathbb{Z}}(-1)^{k}q^{\frac{k(3k+1)}{2}}. $$

How can we prove the above identity? Any suggestions/ideas would be greatly appreciated!

Recall that a $q$-Pochhammer symbol is defined as $$ (x)_n = (x;q)_n := \prod_{l=0}^{n-1}(1-q^l x). $$

I found the following curious $q$-series identity that seems to hold for any $n\geq 0$: $$ (-1)^{n+1}q^{\frac{(n+1)(3n+2)}{2}}\sum_{j\geq 0}q^{j}(q^{j+1})_{n}(q^{j+2n+2})_{\infty} \overset{?}{=} \sum_{\substack{k\in \mathbb{Z}\\ |k| > n}}(-1)^{k}q^{\frac{k(3k-1)}{2}}. $$ Note, the right-hand side is a truncated version of the Euler function $$ \phi(q) := (q)_{\infty} = \sum_{k\in \mathbb{Z}}(-1)^{k}q^{\frac{k(3k-1)}{2}}. $$

How can we prove the above identity? Any suggestions/ideas would be greatly appreciated!