Why is $\sum_{n=1}^{\infty }(-1)^{n+1}\frac{1}{n}.e^{-nx}$ uniformly convergent?

Why is the following series uniformly convergent:$$\sum_{n=1}^{\infty }(-1)^{n+1}\frac{1}{n}.e^{-nx}$$? where $x\geq 0$

I tried the Weierstrass-M test, but it doesn't work here because:$\left | (-1)^{n+1}\frac{1}{n}.e^{-nx} \right |= \frac{1}{n}.e^{-nx}\leq \frac{1}{n}$, and $\sum_{n=1}^{\infty }\frac{1}{n}$ is divergent.

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Why dont you choose $\frac{1}{n(1+nx)}$ to compare to? Plus, for $x=0$, the series is conditionally convergent! – Pedro Tamaroff Apr 8 '12 at 0:18
Unless I'm missing something, this diverges if $x<0$. But I think it converges uniformly on $[0,\infty)$. – Michael Hardy Apr 8 '12 at 0:24
We can use Dirichlet test to deduce the uniform convergence on $[0, \infty)$. But if you want to work with Weierstrass M-test, it suffices to consider the partial sum with the first $2n$ terms. Grouping adjunct two terms, we obtain $$\sum_{k=1}^{n} \frac{e^{-(2k-1)x} + (2k-1)xe^{-(2k-1)x}\left( \frac{1 - e^{-x}}{x} \right)}{(2k-1)(2k)}.$$ Note that the numerator is uniformly bounded by $$1 + \sup_{0 \leq t} \left( t e^{-t} \right) \sup_{0 \leq t} \left( \tfrac{1 - e^{-t}}{t} \right) < \infty$$ on $[0, \infty)$. Thus we obtain uniform convergence by Weierstrass M-test. – Sangchul Lee Apr 8 '12 at 0:30
I agree that $e^{-nx}\leq \frac{1}{n(1+nx)}$, but how does the uniform convergence follow? I can't see it, because you need to prove that $\frac{1}{n(1+nx)}$ is less than equal to $M_{n}$ where $\sum M_{n}$ is convergent. Can anayone provide more details? – M.Krov Apr 8 '12 at 0:30
I don´t know why Didier rejected this but $1/n(1+nx)<1/n^2$ for $x \neq 0$ – Pedro Tamaroff Apr 8 '12 at 16:49

Given $m>n$, for $x\in[0,\infty)$, we have $$\Bigl|\,\sum_{k=n}^m (-1)^{k+1}{1\over k}e^{-kx}\,\Bigr| \le {1\over n}e^{-nx}\le {1\over n}.$$

It follows that $\sum\limits_{n=1}^\infty (-1)^{n+1}{1\over n}e^{-nx}$ is uniformly Cauchy on $[0,\infty)$ and, thus, uniformly convergent on $[0,\infty)$.

Below, are sketched the first few partial sums $S_k=\sum\limits_{n=1}^k (-1)^{n+1}{1\over n} e^{-nx}$ of the series. Note how they "alternate":

More generally, if $(f_n)$ is a decreasing sequence of nonnegative functions that converge uniformly to $0$ on the set $I$, then the series $\sum\limits_{n=1}^\infty (-1)^n f_n$ is uniformly convergent on $I$.

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I was really impressed by your neat answer. Surely it is worth an upvote. Anyway, what graphic software did you use to plot? – Sangchul Lee Apr 8 '12 at 1:24
@sos440 JSXgraph. It's a Javascript library for interactive plotting (but also makes nice diagrams:)). – David Mitra Apr 8 '12 at 1:25
Wow, it's great! Thanks for recommending such a nice one! – Sangchul Lee Apr 8 '12 at 1:29
@M.Krov For fixed $x\ge0$, the sequence $(a_n)$ defined by $a_n={e^{-nx}\over n}$ is decreasing. So, $$|a_k|\ge |a_k-a_{k+1}|\ge |a_k-a_{k+1} +a_{k+2}|\ge\cdots.$$ It's just the argument from the proof that an alternating series converges. – David Mitra Apr 8 '12 at 1:56
@M.Krov Yes, that would work. But the argument I gave is entirely self contained. – David Mitra Apr 8 '12 at 2:28