Recall the notation $(z;q)\_n=(1-z)(1-zq)(1-zq^2)\cdots(1-zq^{n-1})$.


My earlier [MO question](https://mathoverflow.net/questions/420603/congruence-modulo-2-for-q-series) did not find enough interest or yield an answer. Perhaps the modulo $2$ part might have thrown people off. So, I now can write another question which, if proved, would certainly capture the modular problem immediately.


So, I would like to propose:



> 
> **QUESTION.** Are these two $q$-series equal to each other?
> $$\sum\_{n\geq1}\frac{q^{\binom{n+1}2}}{(q;q)\_n}\sum\_{m=1}^n\frac{q^m}{1-q^m}
> =\sum\_{n\geq1}\frac{q^{\frac{n(3n+1)}2}(1+q^{2n+1})(-q;q)\_n}{(q;q)\_n}
> \sum\_{j=1}^n\frac{1+q^{2j}}{1-q^{2j}}.\tag1$$
> 
> 
> 


**Postscript.** This is a response to Vladimir Dotsenko's request. For the [LHS](https://mathoverflow.net/questions/420603/congruence-modulo-2-for-q-series) of my earlier post use (1) above, for the [RHS](https://mathoverflow.net/questions/420603/congruence-modulo-2-for-q-series) use the identity
$$\sum\_{k\geq0}x^{k+2}q^{k+1}\prod\_{j=1}^k(1-xq^j)=\sum\_{n\geq0}(-1)^n\left(x^{3n+2}q^{\frac{(n+1)(3n+2)}2}+x^{3n+3}q^{\frac{(n+1)(3n+4)}2}\right)$$
and replace $x\rightarrow q^{-1}, q\rightarrow q^4$.



Equality of two $q$-series. Proof?
I take it from looking at the previous problem that you are familiar with the [Dyson rank](https://en.wikipedia.org/wiki/Rank_of_a_partition) on partitions with distinct parts. Let's denote by $Q(r,n)$ the number of partitions of $n$ into distinct parts that have rank $r$. The following expression for the generating function of $Q$ is well known:
$$F(x,q)\mathrel{\mathop:}= \sum\_{r,n} Q(r,n)x^rq^n=\sum\_{n\geq 0}\frac{q^{\binom{n+1}{2}}}{(xq;q)\_n}.$$
This comes from grouping the partitions according to the number of parts (or equivalently the largest staircase partition that fits in the diagram).


Another expression from the generating function comes from grouping the partitions according to the largest pentagonal partition that fits in the diagram (the ones that appear as the fixed points of Franklin's involution mentioned by Vladimir Dotsenko in the comments). More specifically this means that for any partition with distinct parts we find the unique $n$ such that the partition either has:


* $n$ parts that are $\geq n+1$ and all other parts are $\le n$
* $n$ parts that are $\geq n+2$, one part $=n+1$ and all other parts are $\le n$


The contribution of the first group to the generating function of $Q$ is given by $\frac{(-x^{-1}q;q)\_n}{(xq;q)\_n}x^nq^{\frac{n(3n+1)}{2}}$ and the contribution of the second group is given by $\frac{(-x^{-1}q;q)\_n}{(xq;q)\_n}x^n q^{\frac{(n+1)(3n+2)}{2}}$. So putting everything together we get
$$F(x,q)=\sum\_{n\geq 0} x^n q^{\frac{n(3n+1)}{2}}(1+q^{2n+1})\frac{(-x^{-1}q;q)\_n}{(xq;q)\_n}.$$
Using these two expressions, you can obtain your identity by evaluating $$\frac{d}{dx}F(x,q)\Bigg\rvert\_{x=1}$$
in two ways.
