For which $\alpha$ does the series $\sum_{n=1}^\infty n^\alpha x^{2n} (1-x)^2$ converges uniformly at $[0,1]$. $$\sum_{n=1}^\infty n^\alpha x^{2n} (1-x)^2$$
Find for which $\alpha$ this series converges uniformly at $[0,1]$.
As $(1-x)^2$ is not dependent of $n$, I thought about rewriting it as:
$$(1-x)^2 \sum_{n=1}^\infty n^\alpha x^{2n} $$
And then you can use the M-test, with $n^\alpha$. Which would make me conclude that it converges if $\alpha < -1$. 
I was thinking that it may also converge at $\alpha \in [-1,1)$. If I take calculate the supremum of $f_n(x)$ I find that this is at $\frac{n}{n+1}$, which gives: $$f_n(\frac{n}{n+1})=n^\alpha (\frac{n}{n+1})^{2n}(1-\frac{n}{n+1})^2= n^\alpha ((1+ \frac{1}{n})^{n})^{-2}(\frac{1}{n+1})^2\leq n^\alpha \frac{1}{(n+1)^2}\leq n^{(\alpha-2)}$$. 
This implies it also converges at $\alpha\in[−1,1)$. The thing I don't understand is that this would mean that the factor $(1-x)^2$ makes it converge at $\alpha\in[-1,1)$. But intuitively I would say that $\sum_{n=1}^\infty n^\alpha x^{2n}$ and $\sum_{n=1}^\infty n^\alpha x^{2n} (1-x)^2$ should converge for the same $\alpha$.
And how could I prove that this series diverges for $\alpha \geq 1$ ?
Edit: If set $\alpha=1$, then I get $$\sum_{n=1}^\infty n x^{2n} (1-x)^2$$
I know the supremum is at $\frac{n}{n+1}$, so this gives $f_n(\frac{n}{n+1})=((1+ \frac{1}{n})^{n})^{-2}\frac{n}{(n+1)^2}$. So if $n\to\infty$, then this goes also to zero. Therefore I would conclude that it converges $\alpha\leq 1$. Is this correct ?

Let's denote by
$$f_n(x)=n^\alpha x^{2n} (1-x)^2.$$
First, we search for which values of $\alpha$ we have the normal convergence of the series that implies the uniform convergence. The supremum of $f_n$ is attained at $x_n=\frac{n}{n+1}$ so 
$$||f_n||_{\infty}=f_n(x_n)=n^\alpha (1+\frac{1}{n})^{-2n}(\frac{1}{n+1})^2\sim e^{-2}n^{\alpha-2},$$
hence the series $\sum_n ||f_n||_{\infty}$ converges for $\alpha<1$. Thus we have the uniform convergence for $\alpha<1.$
Now, what about $\alpha\geq 1$? We have uniform convergence if 
$$\lim_n\sup_{x\in[0,1]}|\sum_{k=n+1}f_k(x)|=0.$$
I'll explain that if the series is of positive terms, then the uniform and normal convergence are same. Indeed, in this case we have
$$\lim_n\sup_{x\in[0,1]}|\sum_{k=n+1}f_k(x)|\geq\lim_n\sum_{k=n+1}f_k(x_k)=\lim_n\sum_{k=n+1}e^{-2}k^{\alpha-2}=+\infty.$$
