I want to show the sequence of functions $f_n : [1,\infty) \to \mathbb{R}$ is uniformly Cauchy. $$ f_n(x) = \sum_{k=0}^{n}\frac{(\ln{x})^k}{k!xe^x} $$
My attempt:
I use Taylor expansion of $e^x \approx 1+ \cdots + \frac{1}{(k-1)!}x^{k-1}$ and $\ln{x}\approx x-1$ but it leads to $$ \Vert f_m - f_n \Vert_{\sup} \leq \sum_{k=n+1}^m \frac{1}{k}.$$ Since RHS does not goes to $0$, it fails. Is there any other good way?
For $x \ge 1$ and $k \ge 2$ is $$ x = e^{\ln x} \ge \frac{(\ln x)^{k-2}}{(k-2)!} $$ and therefore $$ \frac{(\ln{x})^k}{k!xe^x} \le \frac{(\ln x)^2}{(k-1)k e^x} \, . $$ Using $\ln x \le x$ and $e^x > x^2/2$ it follows that $$ \frac{(\ln{x})^k}{k!xe^x} \le \frac{2}{(k-1)k} $$ Since $\sum_{k=2}^\infty \frac{2}{(k-1)k}$ is convergent it follows that $(f_n)$ is uniformly Cauchy.
Concretely: For $2 < n < m$ is $$ \Vert f_m - f_n \Vert_{\sup} \le \sum_{k=n+1}^m \frac{2}{(k-1)k} = 2 \sum_{k=n+1}^m \left( \frac{1}{k-1} - \frac{1}{k} \right) = 2 \left ( \frac 1n- \frac 1m\right) < \frac 2n \, . $$