Uniform Convergence of $\sum_{n=1}^{\infty}x^ne^{-nx}$ on $[0,\infty]$

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.

• Note that with $z:=xe^{-x}$, the series is $\sum_{n=1}^\infty z^n$. – Yves Daoust Nov 24 '20 at 16:42
• $$\sum_{n=1}^\infty \Big( xe^{-x} \Big)^n$$ – Michael Hardy Nov 24 '20 at 18:11
• @user123 Don't vandalize your questions (or, more generally, edit them) after you've received answer. – Gae. S. Nov 25 '20 at 11:51

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 $$-\alpha \leqslant x \leqslant 0$$.
• Note that $\sum_{n=1}^N (-1)^n$ is uniformly bounded and $(|x|e^{|x|})^n \searrow 0$ when $x$ is in the interval $[-\alpha,0]$ – RRL Nov 24 '20 at 18:08
• $\alpha < 1$ from the question, so for $[-\alpha,0]$ that series is convergent?, because (-1)^n is also bounded – user840729 Nov 24 '20 at 18:30