I am able to show that $\lim\limits_{q \rightarrow 1} (1-q)^3 \sum_{n>0} n q^n = 0 $ and $ \lim\limits_{q \rightarrow 1} (1-q)^3 \sum_{n>0} n^2 q^n = 2$ and now I want to prove that
$$ \lim\limits_{q \rightarrow 1} (1-q)^3 \sum_{n>0} n^{1+\epsilon} q^n = 0 $$
for $\epsilon \in (0,1)$. Do you guys have any hints for me doing so? I can't use the methods which I used to proof the first two statements because they are just valid for integral exponents of $n$.
Thanks in advance!
If we recognize that the series in question is a polylogarithm,
$$ \newcommand{\Li}{\operatorname{Li}} \sum_{n=1}^{\infty} n^\alpha q^n = \Li_{-\alpha}(q), $$
then we can take advantage of the alternate representation (specifically the second equation under #2 in the link)
$$ \Li_{-\alpha}(q) = \Gamma(1+\alpha) (-\log q)^{-\alpha-1} + \sum_{n=0}^\infty \frac{\zeta(-\alpha-n)}{n!} \,(\log q)^n $$
which holds for $|\log q| < 2\pi$ and $\alpha \neq -1,-2,-3,\ldots$. As $q \to 1^-$ the series on the right converges to $\zeta(-\alpha)$ and the term on the left blows up (if $\alpha > -1$). Since $\log q \sim q-1$ as $q \to 1$ we may conclude that
$$ \lim_{q \to 1^-} (1-q)^{\alpha+1}\Li_{-\alpha}(q) = \Gamma(1+\alpha) $$
for all $\alpha > -1$.