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

Question

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.

Answer 1

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 $-\alpha \leqslant x \leqslant 0$.