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Prove that

$\sum_{n=1}^{\infty}x^ne^{-nx}$ is uniform convergent on $[0,\infty]$,

I have started by considering the two cases $x\geq0,$ and $x \in [-\alpha,0]$ and applying the M-Test, for $x \geq 0$ I have $\frac{x^n}{e^{nx}}$, but I dont know how to find a series larger than this.

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  • $\begingroup$ Note that with $z:=xe^{-x}$, the series is $\sum_{n=1}^\infty z^n$. $\endgroup$ – Yves Daoust Nov 24 '20 at 16:42
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    $\begingroup$ $$ \sum_{n=1}^\infty \Big( xe^{-x} \Big)^n $$ $\endgroup$ – Michael Hardy Nov 24 '20 at 18:11
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    $\begingroup$ @user123 Don't vandalize your questions (or, more generally, edit them) after you've received answer. $\endgroup$ – Gae. S. Nov 25 '20 at 11:51
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For the case where $x \in [0,\infty)$ show using calculus that $xe^{-x} \leqslant e^{-1}$ and the M-test is applicable to prove uniform convergence since $x^n e^{nx} \leqslant e^{-n}$.

When $x < 0$ write the general term as $x e^{-nx} = (-1)^n (|x|e^{|x|})^n $ and apply the Dirichlet test for uniform convergence. Since $\sum_{n=1}^m (-1)^n$ is uniformly bounded for all $m$, the series converges uniformly if $(|x|e^{|x|})^n \downarrow 0$ monotonically and uniformly as $n \to \infty$.

If $-\alpha \leqslant x \leqslant 0$, then $|x| e^{|x|} \leqslant \alpha e^{\alpha} < 1$ when $\alpha \in (0,\lambda)$ where $\lambda e^{\lambda} = 1$. This ensures the monotonic and uniform convergence condition of the Dirichlet test and we have uniform convergence of $\sum x^n e^{-nx} = \sum (-1)^n (|x|e^{|x|})^n$ for $-\alpha \leqslant x \leqslant 0$.

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  • $\begingroup$ Thank you, (1/e)^n is a geometric series so it converges?, for the other interval can I just do x<0? $\endgroup$ – user840729 Nov 24 '20 at 18:02
  • $\begingroup$ The other case is harder, See hint, $\endgroup$ – RRL Nov 24 '20 at 18:05
  • $\begingroup$ Note that $\sum_{n=1}^N (-1)^n$ is uniformly bounded and $(|x|e^{|x|})^n \searrow 0$ when $x$ is in the interval $[-\alpha,0]$ $\endgroup$ – RRL Nov 24 '20 at 18:08
  • $\begingroup$ $\alpha < 1$ from the question, so for $[-\alpha,0]$ that series is convergent?, because (-1)^n is also bounded $\endgroup$ – user840729 Nov 24 '20 at 18:30
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    $\begingroup$ @user123: You're welcome. The basic test is discussed here in Wikipedia although it does not consider the extension to uniform convergence -- but I showed you that. $\endgroup$ – RRL Nov 24 '20 at 18:46

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