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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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    I don´t know why Didier rejected this but 1/n(1+nx)<1/n^2 for $x \neq 0$2012-04-08

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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":

enter image description here


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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    @M.Krov Yes, that would work. But the argument$I$gave is entirely self contained.2012-04-08