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. 
 A: The series converges uniformly on $[1,\infty)$ by the Weierstrass M-Test (thanks to the exponential term).  To prove it converges uniformly on $[0,1]$, use properties of convergent alternating series.  For any convergent alternating series of reals with terms nonincreasing in absolute value, the absolute value of the difference between the $n$th partial sum $s_n$ and the sum $s$ is less than or equal to the absolute value of the $n+1$st term.
By the way, the series is an alternating series for all $x \geq 0$.
The proof above for $0\leq x \leq 1$ looks like it works for all $x \geq 0$, but I'm going to stick with what I wrote above.
A: The $n$th partial sum can be written explicitly for every $x>0.$ (think $\sum_{k=1}^n a^k/k$). Should be easy from then on.
A: 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 $0$
