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\*}



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> **QUESTION.** Is there a similar expression for the following?
> $$\sum\_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(1-q^n)^m\,(q;q)\_n}.$$
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**Remark.** Let's at least try out this for small values of $m$, say $m=3$ or $m=4$.



Is there a generalization of these q-series identities?
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
$$H\_1+H\_2=\sum\_{n\geq 1}\frac{q^n}{1-q^n}+\frac{1}{2}\sum\_{n\geq 1}\left(\frac{q^n}{1-q^n}\right)^2+\frac{1}{2}\left(\sum\_{n\geq 1}\frac{q^n}{1-q^n}\right)^2$$
$$=\frac12\sum\_{n\geq 1}\frac{(n+1)q^n}{1-q^n}+\frac12\left(\sum\_{n\geq 1}\frac{q^n}{1-q^n}\right)^2.$$



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