I want to show uniform convergence of $\sum_{k=1}^{\infty} x^k e^{-kx}$ where $x \in [0,\infty)$. I am not sure how to demonstrate this since any avenues I have pursued have not worked. Does the Weierstrass M-test work for this, or should I be using something else?
I know the M-test works for just the $e^{-kx}$ portion, but I can't figure out how this will work for the whole.
You should be able to proceed by showing the function $f(x) = x^k*e^{-kx}$ is bounded. This is the M-test for uniform continuity.
So, we want to show there exists $M>0$ such that $|x^k*e^{-kx}| < M$ for all $x \in [0, \infty]$ and all $k \in [1, \infty]$. For finite $x$, this seems plausible. Let's first consider what happens as $x \rightarrow \infty$.
$\displaystyle \lim_{x \rightarrow \infty}x^k*e^{-kx} $ = $\displaystyle \lim_{x \rightarrow \infty}\frac{x^k}{e^{kx}}$.
Now we can use L'Hopital's rule $k$ times, and we're left with $\displaystyle \lim_{x \rightarrow \infty}\frac{1}{e^{x}} = 0$.
It ended up not even mattering what happened to $k$ in that limit.
What matters is the function values become very small as x becomes very large, so even as we sum k to infinity, we're ultimately shrinking our finite sums down, so the infinite sum converges.