Not an answer, but a research idea: first, replace $x$ by $k$ then write your sum as follows:

$\sum_{k=1}^n \frac{k^a}{1-q^k} = \sum_{k=1}^\infty \left(c_k \frac{k^a}{q^k}\right) \frac{q^k}{1-q^k}$

where $c_k=0$ for $k>n$ and 1 otherwise. This is a Lambert series; see wikipedia. That is, let

$a_k = \left(c_k \frac{k^a}{q^k}\right)$

and then apply a Dirichlet convolution (see wikipedia) to get

$b_m = (a*1)(m) = \sum_{k|m} a_k$

and thus

$\sum_{k=1}^n \frac{k^a}{1-q^k} = \sum_{m=1}^\infty b_m q^m$

So now you have an infinite sum, in place of a finite one. A different, less fancy way to say what I just said is to just write

$\frac{1}{1-q^k} = \sum_{m=0}^\infty (q^k)^m$

and just plug that in, and then exchange order of summation. That way, you don't have a pesky denominator; you traded it for an infinite sum, which is pesky in a different way. (you might also find more willing help on math exchange instead of mathoverflow?)

Not an answer, but a research idea: first, replace $x$ by $k$ then write your sum as follows:

$\sum_{k=1}^n \frac{k^a}{1-q^k} = \sum_{k=1}^\infty \left(c_k \frac{k^a}{q^k}\right) \frac{q^k}{1-q^k}$

where $c_k=0$ for $k>n$ and 1 otherwise. This is a Lambert series; see wikipedia. That is, let

$a_k = \left(c_k \frac{k^a}{q^k}\right)$

and then apply a Dirichlet convolution (see wikipedia) to get

$b_m = (a*1)(m) = \sum_{k|m} a_k$

and thus

$\sum_{k=1}^n \frac{k^a}{1-q^k} = \sum_{m=1}^\infty b_m q^m$

So now you have an infinite sum, in place of a finite one. A different, less fancy way to say what I just said is to just write

$\frac{1}{1-q^k} = \sum_{m=0}^\infty (q^k)^m$

and just plug that in, and then exchange order of summation. That way, you don't have a pesky denominator; you traded it for an infinite sum, which is pesky in a different way.

I suspect my suggestion above is completely useless; I cant quite figure out what you want. You might also find more willing help on math exchange instead of mathoverflow?