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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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    Why dont you choose $\frac{1}{n(1+nx)}$ to compare to? Plus, for $x=0$, the series is conditionally convergent!2012-04-08
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    Unless I'm missing something, this diverges if $x<0$. But I think it converges uniformly on $[0,\infty)$.2012-04-08
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    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.2012-04-08
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    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?2012-04-08
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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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    I was really impressed by your neat answer. Surely it is worth an upvote. Anyway, what graphic software did you use to plot?2012-04-08
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    @sos440 [JSXgraph](http://www.jsxgraph.org/). It's a Javascript library for interactive plotting (but also makes nice diagrams:)).2012-04-08
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    Wow, it's great! Thanks for recommending such a nice one!2012-04-08
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    @David Mitra: How did you get this inequality: $ \Bigl|\,\sum_{k=n}^m (-1)^{k+1}{1\over k}e^{-kx}\,\Bigr| \le {1\over n}e^{-nx}\le {1\over n}. $? Shouldn't be: $\leqslant \frac{1}{n}e^{-nx}+\frac{1}{n+1}e^{-(n+1)x}+...+\frac{1}{m}e^{-mx}$?2012-04-08
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    @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](http://en.wikipedia.org/wiki/Alternating_series).2012-04-08
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    @David Mitra: That's right. Thanks. Also, is there a difference between a sequence being Cauchy and Uniformly Cauchy?2012-04-08
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    @M.Krov Yes, similar to bieng "convergent" and "uniformly convergent". Note the term on the right in the inequality in the answer is independent of $x$.2012-04-08
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    @David Mitra: Thanks a lot for this useful information.2012-04-08
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    @M.Krov You're welcome; glad to help.2012-04-08
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    @ David Mitra: Based on the theorem you gave:"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 $I$", I can do the following: 1)- prove that $\frac{1}{n}e^{-nx}$ is a decreasing sequence of functions. 2)- Prove that: $lim_{n\rightarrow \infty }sup\left \{ \left | f_{n(x)-f(x)} \right | \right \}=0$ whcih means that the sequence of functions is uniformly convergent to $f(x)=0$ and the uniform convergence of the series follows.2012-04-08
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    @ David Mitra: Is my conclusion above correct?2012-04-08
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    @M.Krov Yes, that would work. But the argument I gave is entirely self contained.2012-04-08