If $\alpha \geqslant 1$, the sequence $(m_{n})_{n \geqslant 1}$ defined by $$m_{n}=\sup _{x\in[0 ; 1]}|U_{n}(x)|=\left|u_{n}\left(\frac{n}{n+1}\right)\right|=\frac{n \alpha}{n+1} \cdot\left(1+\frac{1}{n}\right)^{-n}$$ diverges: $(u_{n})_{n\geqslant 1}$ does not converge uniformly on $[0,1]$. Thus,$\sum_{n \geqslant 1} u_{n}$ has not uniform convergence on $[0,1]$ for $\alpha \geqslant 1$. If $\alpha<0$, $\sum_{n \geq 1} u_{n}$ normally converges on $[0,1]$. Thus $\sum_{n \geqslant 1} u_{n}$ uniformly converges on $[0,1]$.
Let $0 \leqslant \alpha<1$. We introduce the sequence $(R_{n})_{n\geqslant 1}$ defined by \begin{align*} R_{n}:[0, 1] & \longrightarrow \mathbb{R} \\ x & \longmapsto R_{n}(x)=\sum_{k=n+1}^{+\infty} k^{\alpha} \cdot x^{k} \cdot(1-x) \end{align*}
For all $n \in \mathbb{N}^{*}$, \begin{align*} k \geqslant n+1 & \Rightarrow k^{\alpha} \geqslant(n+1)^{\alpha}\\ & \Rightarrow k^{\alpha} \cdot x^{k} \cdot(1-x) \geqslant(n+1)^{\alpha} \cdot x^{k}(1-x)\\ & \Rightarrow R_{n}(x) \geqslant \sum_{k=n+1}^{+\infty}(n+1)^{\alpha} \cdot x^{k}(1-x). \end{align*} As $\sum\limits_{k=n+1}^{+\infty}(n+1)^{\alpha} \cdot x^{k} \cdot(1-x)=(n+1)^{\alpha} \cdot x^{n+1}$, then for all $0<a<1$, \begin{align*} x \geqslant a & \Rightarrow R_{n}(x) \geqslant(n+1)^{\alpha} \cdot a^{n+1}\\ & \Rightarrow R_{n}(x) \geqslant a^{n+1}>0, \end{align*} we deduce from this inequality that $\sum_{n \geqslant 1} u_{n}$ does not converge uniformly on $[a,1]$.
Now for $x\in[0,a)$, I don't really see how to solve the problem. Is it really necessary to study this case or we could just try to see what happens when $x=0$?
Suppose $0 \leqslant \alpha < 1$ and $x \in [a,1]$ with $0< a < 1$. We have
$$\left|\sum_{k = n+1}^\infty k^\alpha x^k(1-x)\right| \geqslant n^\alpha (1-x) \sum_{k=n+1}^\infty x^k = n^\alpha x^{n+1},$$
and with $x_n = 1 - 1/n \in [a,1]$ for all sufficiently large $n$,
$$\sup_{x \in [a,1]}\left|\sum_{k = n+1}^\infty k^\alpha x^k(1-x)\right| > n^\alpha (1-1/n)^{n+1} \underset{n \to \infty}\longrightarrow \begin{cases}\infty, & 0 < \alpha < 1\\ e^{-1}, &\alpha = 0 \end{cases}$$
Hence, the convergence is not uniform on $[a,1]$.
Alternatively,
$$\sum_{k=n+1} ^{2n} k^\alpha x^k (1-x) \geqslant n \cdot n^\alpha\cdot x^{2n}\cdot (1-x),$$
and with $x_n = 1 - 1/n$ we have
$$\sup_{x \in [a,1]}\left|\sum_{k = n+1}^\infty k^\alpha x^k(1-x)\right| \geqslant \sup_{x \in [a,1]}\sum_{k=n+1} ^{2n} k^\alpha x^k (1-x)> n\cdot n^\alpha \cdot (1- 1/n)^{2n}\cdot (1 - (1-1/n)) \\=n^\alpha (1-1/n)^{2n} \underset{n \to \infty}{\not\to} 0$$