$$ f_n(x)=\frac{n^\alpha x}{e^{nx}} $$ I'm struggling with the idea of uniform convergence, so i'm not sure this solution is correct.
The function is defined between $[0,1]$, positive and continuous. Also $\displaystyle\lim_{n\to\infty} f_n(x)=0$ for every $x\in[0,1]$.
I check for the maximum of the function: $$ (\frac{n^\alpha x}{e^{nx}})'= \frac{n^\alpha e^{nx}-ne^{nx}n^\alpha x}{(e^{nx})^2}=0 $$ $$ n^\alpha e^{nx}=ne^{nx}n^\alpha x \iff nx=1 \iff x=\frac{1}{n} $$ And when we put $\frac{1}{n}$ in the original function: $$ \frac{n^\alpha \frac{1}{n}}{e}=\frac{n^{\alpha-1} }{e} $$ And when $n\to\infty$ the maximum goes to infinity for every $\alpha > 1$ so there is no uniform convergence.
When $\alpha = 1$ The maximum of the function is $\frac{1}{e}$ for every n and so $|f_n(x)-f(x)|\not<\epsilon$ for $\epsilon < \frac{1}{e} \implies$ no uniform convergence.
When $0<\alpha<1$ the maximum of the function is $\frac{\frac{1}{n^{1-\alpha}}}{e}$ and when $n\to\infty$ it equals $0$ so $|f_n(x)-f(x)|<\epsilon$ for every $x$ $\implies$ there's uniform convergence for $0 < \alpha < 1$
When $\alpha=0$ the maximum of the function is simply $f_n(x)=\frac{n^0-1}{e}=0$ and thus there's uniform convergence (because if the max is zero than that's true for every other x because this function is positive).
When $\alpha<0$ the maximum of the function is $\frac{n^{\alpha - 1}}{e}=\frac{\frac{1}{n^{-\alpha + 1}}}{e}$.
The $n$ is to the power of a positive number so when $n\to\infty$ the maximum approaches zero $\implies$ there's uniform convergence when $\alpha<0$
So to summarize: the function series has uniform convergence when $\alpha<1$.