Determine the sequence of functions $f_n(x)=nx^2e^{-nx}$ is uniformly convergent or not. Let $\{f_n\}^{+∞}_{n=1}$ be a sequence of functions where $f_n(x)=nx^2e^{-nx}$ where $x$ belongs to $[1;+\infty)$ and $n$ belongs to $\mathbb N$ (all natural numbers).
Determine whether the sequences of functions ${f_n}$ is uniformly convergent or not.
I computed $g_n(x)=|f_n(x)−f(x)|=nx^2e^{-nx}$ then solved the equation $g'_n(x)=0$, I get $x=\frac{2}{n}$ or $x=0$. But $0$ and $\frac{2}{n}$ do not belong to $[1;+\infty)$, for all $n \in \mathbb{N}$.
So what I can do to find the supremum of $|f_n(x)−f(x)|$?

For each $x\in[1,\infty)$, $\lim_{n\to\infty}f_n(x)=0$. In other words, $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function. On the other hand, if $n\geqslant2$, then$$\bigl(\forall x\in[1,\infty)\bigr):f_n'(x)=-e^{-nx}nx(nx-2)\leqslant0,$$and therefore $f_n$ is decreasing. Since $f_n$ is also non-negative,\begin{align}\sup_{x\in[1,\infty)}|f_n(x)-0|&=\sup_{x\in[1,\infty)}f_n(x)\\&=f_n(1)\\&=\frac n{e^n}.\end{align}Since $\lim_{n\to\infty}\sup_{x\in[1,\infty)}|f_n(x)-0|=0$, the convergence is uniform.
