Trouble with proving uniform convergence prove : $$\sum_{n=1}^{\infty } \dfrac{nx}{e^{nx}}$$
converge for each x $\in R$ , converge uniformly in $[\alpha,\infty)$ for $\alpha >0$ and doesn't converger uniformly at $[0,\infty)$. 
Having trouble with that question, 
Thanks for helping.

Here, we show that the series $\sum_{n=1}^\infty nxe^{-nx}$ fails to be uniformly convergent for $x\in [0,\infty)$.
Take $\epsilon=e^{-1}$.  Then, for $x=1/N$ we have
$$\begin{align}
\left|\sum_{n=1}^\infty nxe^{-nx}- \sum_{n=1}^{N-1} nxe^{-nx}\right|&=\sum_{n=N}^\infty (n/N)e^{-n/N}\\\\
&\ge \sum_{n=N}^\infty e^{-n/N}\\\\
&=\frac{e^{-1}}{1-e^{-1/N}}\\\\
&>\frac{1}{e}\\\\
&=\epsilon
\end{align}$$
Therefore, there exists a number $\epsilon =e^{-1}>0$, such that for all $N>1$ there exists a number $x=1/N$, such that $$\left|\sum_{n=1}^\infty nxe^{-nx}- \sum_{n=1}^{N-1} nxe^{-nx}\right|>\epsilon$$
And we are done!
