Is there a generalization of these q-series identities?

Question

Denote $(q;q)_n=(1-q)(1-q^2)\cdots(1-q^n)$.

The below three identities are known. \begin{align*} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(q;q)_n} &=1-\sum_{n\in\mathbb{Z}}(-1)^nq^{\frac{n(3n+1)}2}, \\ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(1-q^n)\,(q;q)_n} &=\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}, \\ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(1-q^n)^2\,(q;q)_n} &=\frac12\sum_{n=1}^{\infty}\frac{(n+1)q^n}{1-q^n}+\frac12\left(\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}\right)^2. \end{align*}

QUESTION. Is there a similar expression for the following? $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(1-q^n)^m\,(q;q)_n}.$$

Remark. Let's at least try out this for small values of $m$, say $m=3$ or $m=4$.

Answer 1

There are many identities in the literature that express these sort of $q$-hypergeometric sums in terms of Lambert series. The relevant ones here are the ones by Dilcher [1]: $$\sum_{n\geq 1}\frac{(-1)^{n-1}q^{\binom{n+1}{2}+(m-1)n}}{(1-q^n)^{m}(q;q)_n}=\sum_{1\le n_1\le n_2\le \cdots \le n_m}\frac{q^{n_1+n_2+\cdots+n_m}}{(1-q^{n_1})(1-q^{n_2})\cdots (1-q^{n_m})}$$ If we denote the right hand side by $H_m$, we can transform your sum as follows $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(1-q^n)^m\,(q;q)_n}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}(q^n+1-q^n)^{m-1}}{(1-q^n)^m\,(q;q)_n}$$ $$=\sum_{k=0}^{m-1}\binom{m-1}{k}H_{m-k} \tag{*}$$ Notice that each $H_k$ is just the $k$-th complete homogeneous symmetric polynomial on the infinite set of variables $\{\frac{q}{1-q}, \frac{q^2}{1-q^2}, \dots\}$. I find $(*)$ to be one of the neatest ways of expressing the right hand side, but if you prefer you can rearrange the terms to give you equivalent expressions. For example, for $m=2$ we get $$H_1+H_2=\sum_{n\geq 1}\frac{q^n}{1-q^n}+\frac{1}{2}\sum_{n\geq 1}\left(\frac{q^n}{1-q^n}\right)^2+\frac{1}{2}\left(\sum_{n\geq 1}\frac{q^n}{1-q^n}\right)^2$$ $$=\frac12\sum_{n\geq 1}\frac{(n+1)q^n}{1-q^n}+\frac12\left(\sum_{n\geq 1}\frac{q^n}{1-q^n}\right)^2.$$

[1] Dilcher, K.: Some q-series identities related to divisor functions. Discrete Math. 145, 83–93 (1995)