Uniform convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^n}{ne^{nx}}$ Prove that
$$ f(x) = \sum\limits_{n=1}^\infty \frac{(-1)^n}{ne^{nx}} = \sum\limits_{n=1}^\infty \frac{(-e^{-x})^n}{n}$$
is uniform convergent for $x \in [0,\infty)$.
Attempt:
At first, this looked a lot like the alternating series. At $x=0$, it is! And for $ x > 0$, $f (x)$ appears to satisfy $|f(x)| < \frac{1}{n}$, but I haven't been able to strengthen this assumption in any way. As the sum of $\frac{1}{n}$ diverges, this isn't useful to apply the M-test for uniform convergence.
Next, I've simply tried to find a way to prove pointwise convergence to $0$ for $f(x)$, but haven't been successful. If it could be shown that $f(x)$ has a pointwise limit function over an arbitrary closed interval $[a,b], 0 \le a$, then Dini's Theorem gives uniform convergence. But, I haven't been able to reach that conclusion. 

First we show that it converges in that range, and now we can prove the uniform convergence using the inequality Laibniz:
$|\sum\limits_{n=1}^\infty \frac{(-1)^n}{ne^{nx}}-\sum\limits_{n=1}^N\frac{(-1)^n}{ne^{nx}}|<\frac{1}{(N+1)e^{(N+1)x}}$       that approaches $0$
