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 $|f_n(x)−f(x)|$?
 A: 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.
A: Note that $e^{nx} =  1 + nx + \frac{(nx)^2}{2} + \cdots > \frac 12 n^2x^2$ for all $x > 0$. So we have $f_n(x) = nx^2 e^{-nx} < \frac 2n$. Since $2/n$ converges to $0$ we must have pointwise limit function $f(x) = 0$. So $\vert f_n(x) -f(x) \vert < \frac 2n$. So supremum must be less that $2/n$ for each $n \in \mathbb{N}$.
Can you finish it now?
A: By noting that it's derivative vanishes at $u=2$, we get that for $u\ge0$,
$$
u^2e^{-u}\le\frac4{e^2}\tag1
$$
The pointwise limit of the functions is $0$. To show uniform convergence, we need to show that $\lim\limits_{n\to\infty}\|f_n-0\|_\infty=0$. Using $(1)$, we have
$$
\begin{align}
nx^2e^{-nx}
&=\frac1n\,\overbrace{(nx)^2e^{-nx}\vphantom{\frac1n}}^{\le4/e^2}\\
&\le\frac4{ne^2}\tag2
\end{align}
$$

$(2)$ actually shows that the sequence of functions converges uniformly on $[0,\infty)$. Convergence on $[1,\infty)$ is a bit simpler since the sequence $f_n(x)$ is non-increasing when $nx\ge1$. Thus, for $x\in[1,\infty)$, $f_n(x)$ is non-increasing for $n\ge1$.
