Show $\sum_{k = 1}^{\infty}x^{k}e^{-kx}$ converges uniformly in $[0,A]$

Question

Show $\sum_{k = 1}^{\infty}x^{k}e^{-kx}$ converges uniformly in $[0,A]$

Attempt

There are two ways I've thought about how to approach the problem. One was to show that the series is Cauchy and the other was to apply the Weirstrauss M-Test. In both cases I have to get some sort of bound on my series.

Using the Cauchy approach:

Letting $S_{n}(x) = \sum_{k = 1}^{n} x^{k}e^{-kx}$

$$\|S_{n}(x) - S_{n+m}(x)\|_{\infty} = \sup_{x \in [0,A]} \Bigg|\sum_{k = 1}^{n} x^{k}e^{-kx} - \sum_{k = 1}^{n+m} x^{k}e^{-kx} \Bigg| \leq \sup_{x \in [0,A]} \sum_{k = n}^{n+m} \frac{|x|^{k}}{|e^{x}|^k}$$

So I want to argue in some way that this last expression $\frac{|x|^{k}}{|e^{x}|^k}$ can bound my series. I know for if I put in any $x \in [0,A]$ the ratio of the terms will be less than 1 and from there argue that this is a geometric series. But as can be observed I have an issue at $x = 0$ because the string of inequalities won't be true. It is similar to a fraction of the form $\frac{x^k}{k!}$ which if it was the expression I was working with I could simply compare the terms as $\frac{x^k}{k!} \leq \frac{A^k}{k!}$ and that would work.

Suggestions or comments on whether either of my approaches is correct?

Answer 1

Hint: Try Weierstrass after answering this: What is the maximum of $x^ke^{-kx}$ on $[0,\infty)$?

Answer 2

It is well known (and easily seen) that $\sum_ky^k$ converges uniformly for $|y|\le a$ for any $a < 1$. Noting that $y = xe^{-x} \le 1/e < 1$ for all $x\in [0,\infty)$ shows that your series converges uniformly on $[0,\infty)$.

Or directly ($x\ge 0$): $$ \left|\sum_{k=1}^\infty y^k - \sum_{k=1}^ny^k\right| = \left|\frac 1{1-y} - \frac{1-y^{n+1}}{1-y}\right| = \frac{x}{e^x-x}\cdot\left(\frac{x}{e^x}\right)^n\le\frac 1{e-1}\cdot\left(\frac 1e\right)^n. $$