Prove that $f_n(x) = \left(\frac{x}{n}\right)^ne^{-x}$ converges uniformly

Question

Prove that $f_n(x) = \left(\frac{x}{n}\right)^ne^{-x}$ converges uniformly at $[1,\infty)$.

So for every $x$, there's $N\in\mathbb{N}$, such that for all $n>N$: $\frac{x}{n} < 1$. Therefore, easy to see that for every $x$: $$\lim_{n\to\infty} f_n(x) = 0$$

Hence, we can tell that $f_n(x)$ converges pointwise to $0$.

Now, I've looked at $f_n'(x)$ and compared it to $0$ looking for a maximum point. I've found that when $x=n$, $f_n'(x)$ has a maximum.

So, $$\lim_{n\to\infty} f_n(n) = e^{-x} \ne 0$$.

Did I just prove $f_n(x)$ doesn't converge uniformly on $[1,\infty)$ (because according the answer, it actually does).

Answer 1

Note $f_n(n) = e^{-n}$, not $e^{-x}$.

Answer 2

Here is an inequality-packed way to prove uniform convergence. First use $$\left(\frac{x}{n} \right)^n = \frac{x^n}{n^n} \leq \frac{1}{n}\cdot \frac{x^n}{n!} \leq \frac{1}{n} \left(\sum_{k=0}^\infty \frac{x^k}{k!}\right) = \frac{e^x}{n}$$ to obtain $$\left(\frac{x}{n} \right)^n \leq \frac{e^x}{n}$$ which implies $$\left(\frac{x}{n} \right)^ne^{-x} \leq \frac{1}{n}$$ for all $x \in [0, \infty)$ and $n \in \Bbb{N}\setminus \{ 0\}$. Now let $\varepsilon>0$. Then there is a natural number $N \in \Bbb{N}$ such that $$\varepsilon> \frac{1}{N} \geq \left(\frac{x}{N} \right)^Ne^{-x}$$ which establishes uniform convergence of $f_n$.