Uniform convergence of $f_n(x)=\frac{x^{\alpha}}{1+nx}$ While studying real analysis I came across the following:
If $0<\alpha<1$, $x \in (0,1)$ and $n=1,2,3,...$ for what values of $\alpha$ does $f_n(x)=\frac{x^{\alpha}}{1+nx}$ converge uniformly?
Before, I had to prove that $f_n(x)=\frac{1}{1+nx}$ doesn't converge uniformly whereas $f_n(x)=\frac{x}{1+nx}$ does converge uniformly.
I somehow could manage to prove, using C-S inequality, that $f_n(x)$ converges uniformly for $\alpha\geq \frac12$, but have no clue what to do next.
Any help would be highly appreciated. Thanks and regards.

Note that $f_n(x)$ has a maximum at $x_n = \alpha/[n(1-\alpha)]$.  For fixed $\alpha \in(0,1)$ and $n$ sufficiently large we have $x_n \in(0,1)$.
$$\lim_{n \to \infty}\sup_{x \in (0,1)}|f_n(x)| = \lim_{n \to \infty}f_n(x_n) = \lim_{n \to \infty}(1-\alpha)\left(\frac{\alpha}{1-\alpha}\right)^\alpha n^{-\alpha} = 0.$$
Hence, the convergence is uniform on $(0,1)$.
