I have the problem with the uniform convergence of:
$$\ \sum_{n=0}^\infty \frac{xe^{-nx}}{e^x-1} ~~~~ x\in (0, \infty)$$
I have checked $\ lim_{x\to0^+} = 0 $, however I have a problem with $\ x\to \infty $. After d'Hospital I got a little curious results.
Later I would like to tst the almost uniform convergnce and pointwise convergence, however, I can't move on without this limit. I have to compare the sup of the function with this limit, if it tends to zero.
(I suppose I would need a little help with almost uniform convergence)
Pointwise convergence follows from the fact that $\sum e^{-nx} <\infty$ for each $x \in (0,\infty)$.
For $x \geq a$ with $a >0$ note that $\frac x {e^{x}-1} $ is bounded and $e^{-nx} \leq e^{-an}$. Hence uniform convergence holds on $[a, \infty)$. Hence the series converges almost uniformly. Uniform convergence fails near $0$: $\frac {xe^{-nx}} {e^{x}-1}$ does not tend to $0$ uniformly as seen by taking $x=\frac 1n$.