From recursive polynomials to a $q$-series Fix an integer $k$. Let $b_n(q)$ be the polynomial defined by the recursive equation
$$b_n(q)=\binom{n+k-1}k+(1+q^{n-1})b_{n-1}(q), \qquad n\geq1,$$
initializing with $b_0(q)=0$. 
I run into the below through experiment.

QUESTION. Denote $(q)_m=(1-q)(1-q^2)\cdots(1-q^m)$. Is this true?
  $$\lim_{n\rightarrow\infty}\left(b_n(q)-\binom{n+k}{k+1}\right)
=\sum_{m=1}^{\infty}\frac{q^{\binom{m+1}2}}{(q)_m(1-q^m)^{k+1}}.$$

NOTE. The case $k=-1$ recovers the generating function for partitions into distinct parts.
 A: Let $a_n(q)=b_n(q)-\binom{n+k}{k+1}$ and $a_\infty(q)=\lim_{n\to\infty} a_n(q)$. Then
$$\begin{align}a_n(q)&=\binom{n+k-1}{k}-\binom{n+k}{k+1}+(1+q^{n-1})\left(a_{n-1}(q)+\binom{n+k-1}{k+1}\right)\\&=q^{n-1}\binom{n+k-1}{k+1}+(1+q^{n-1})a_{n-1}(q).\end{align}$$
By following each $\binom{n+k-1}{k+1}$ term,
$$a_\infty(q)=\sum_{n\ge 0}\binom{n+k-1}{k+1}q^{n-1}(-q^n)_\infty=\sum_{l\ge 0}\binom{l+k}{k+1}q^l(-q^{l+1})_\infty.$$
On the other side, Taylor expand $(1-q^m)^{-k-1}$, use the sum representation of $(-q^{n+1})_\infty$, and apply the hockey stick identity to get
$$\begin{align}
\sum_{m\ge 1}\frac{q^{\binom{m+1}{2}}}{(q)_m(1-q^m)^{k+1}}&=\sum_{m\ge 1}\frac{q^{\binom{m}{2}+m}}{(q)_m}\sum_{n\ge 0}\binom{n+k}{k}q^{mn}\\
&=\sum_{n\ge 0}\binom{n+k}{k}\sum_{m\ge 1}\frac{q^{\binom{m}{2}}}{(q)_m}q^{(n+1)m}\\
&=\sum_{n\ge 0}\binom{n+k}{k}((-q^{n+1})_\infty-1)\\
&=\sum_{n\ge 0}\binom{n+k}{k}\sum_{i\ge 1}q^{n+i}(-q^{n+i+1})_\infty\\
&=\sum_{l\ge 1}q^{l}(-q^{l+1})_\infty\sum_{n=0}^{l-1}\binom{n+k}{k}\\
&=\sum_{l\ge 1}q^{l}(-q^{l+1})_\infty\binom{l+k}{k+1}.
\end{align}$$
