Find for which values of $\alpha\in\mathbb R, f_n(x)$ converge uniformally

Question

The question is

"Find for which values of $\alpha\in\mathbb R$ such that $ f_n(x)$ converge uniformally in $[0,\infty)$ where $f_n(x)=n^\alpha x \dot e^{-nx} $".

For $\alpha<0$, the sequence converges uniformally to $0$ (since $\forall x\in[0,\infty):e^{nx}>x $ hence the denominator is bigger then the counter and tend to infinity). My problem is I don't know how to procedd further... The book claims the sequence converges $\forall \alpha<1$. Why is this answer correct?

Answer 1

We know that the $f_n$ are continuous and non-negative, that $f_n(0) = 0$ and $\lim_{x\to\infty} f_n(x) = 0$, since the exponential function grows much faster than $x$.

So let's try to find $\max \{ f_n(x) \colon x \in [0,\,\infty)\}$. That must be a critical point of $f_n$, so let's try to find the zeros of $f_n'$.

$$f_n'(x) = n^\alpha e^{-nx} + n^\alpha x \cdot(-n)e^{-nx} = n^\alpha e^{-nx} (1-nx).$$

The only zero of $f_n'$ is therefore $x = \frac1n$, and thus that must be the point where $f_n$ attains its maximum.

$$f_n(\frac1n) = n^\alpha \frac1n e^{-n\frac1n} = n^{\alpha-1}/e,$$

and $n^{\alpha-1} \to 0 \iff \alpha < 1$.

Answer 2

It's clear that the sequence $(f_n)$ is pointwise convergent to the zero function on the interval $[0,+\infty)$.

We have $$f'_n(x)=n^\alpha e^{-nx}(1-nx)=0\iff x=\frac{1}{n}=x_n$$ hence we find $$||f_n||_\infty=f_n(x_n)=n^{\alpha-1}e^{-1}\to0\iff\alpha-1<0\iff\alpha<1$$ and we conclude that the sequence is uniformly convergent if $\alpha<1$