Let $0< \alpha <1$ and $q>1.$ 
Consider the (alternating) series: $$
\sum\_{k=1}^\infty 
(-1)^k \frac{q^k (q^k-1)^\alpha}{(q^k-1)\dots (q-1)}.$$
Denote its sum by $f(q,\alpha).$ 


Prove (or disprove) that $f(q,\alpha)\neq 0$ for all
$\alpha\in(0,1)$ and all $q> 1.$


Computer computation confirms that there are no zeros for $ 1<q<2$ and $0<\alpha<1.$
For $\alpha\in N\_0,$ we obtain 0 by Euler's formula. (Added on September 5 after the discussion, thanks to all who participated).


Remark. For $0<\alpha <1/2,$ and $q\geq 2$ the proof is rather simple, but, for $\alpha$ close to 1,I have not succeeded. The problem is related to the study of the convergence for the $q$-Bernstein polynomials.



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.


