The sum of a series

I have no doubt that you have figured this exercise out by now but let me post the solution just for the fun of it.

Denote $U_k=\prod_{m=1}^k\frac 1{q^m-1}$ with the usual convention $U_0=1$.

We have $\prod_{k\ge 1}(1-q^{-k}y)=\sum_{k\ge 0}(-1)^kU_ky^k$. In particular, it implies that $$ \sum_{k\ge 1}(-1)^k U_{k-1}y^k<0\text{ for all } 0<y<q $$ The sum you are interested in is $$ \sum_{k\ge 1}(-1)^k U_{k-1}\frac{q^k}{(q^k-1)^\beta} $$ where $\beta=1-\alpha$.

Now just write $$ \frac{q^k}{(q^k-1)^\beta}=(q^{1-\beta})^k(1-q^{-k})^{-\beta}=(q^{1-\beta})^k\sum_{m\ge 0}c_{m}(\beta)q^{-mk}=\sum_{m\ge 0}c_m(\beta)(q^{1-m-\beta})^k $$ and observe that all $c_m(\beta)$ (the Taylor coefficients of $x\mapsto (1-x)^{-\beta}$ at $x=0$) are positive.