Pointwise and uniform convergence of series 
Consider the series $\displaystyle \sum_{n\geq 1}n^\alpha x^n(1-x^2)$ for all $\alpha\in\mathbb{R}$.

*

*Prove that the series converges pointwise on $[-1,1]$ for all
$\alpha\in\mathbb{R}$.


*Let $\delta>0$. Prove that for all $\alpha\in\mathbb{R}$ the series converges uniformly on every interval $[-1+\delta,1-\delta]$.


*For which $\alpha$ do we have uniform convergence on $[-1,1]$?

(1): For $x=\pm 1$, the series is equal to zero. Let $a_n=n^\alpha x^n(1-x^2)$, then $|a_n|^{1/n}=(n^{1/n})^\alpha |x||1-x^2|^{1/n}\xrightarrow{n\to\infty}|x|$, by the root test the series converges for $|x|<1$. Is this correct?
(2): I try to apply the Weierstrass M-test, $|n^\alpha x^n(1-x^2)|\leq |n^\alpha x^n|$ but I can't seem to get rid of the $n^\alpha$ term? I also tried to prove that $\lim_{n\to\infty}\sup_{x\in [-1+\delta,1-\delta]}|n^\alpha x^n(1-x^2)|=0$, but differentiating gives a mess and this mess doesn't seem to go to zero. Could I get a hint?
 A: $(a)$ is correct.
For $(b)$ you have $x \in [-1+\delta, 1-\delta]$ which implies $|x| \le |1-\delta|^n$.
Therefore:
$$\left| n^\alpha x^n \underbrace{(1-x^2)}_{\le 1}\right| \le n^\alpha |1-\delta|^n$$
Just like you did in $(a)$, the root test gives that $\sum_{n=1}^\infty n^\alpha |1-\delta|^n < +\infty$ since $|1-\delta| \in [0,1\rangle$. The Weierstrass $M$-test gives uniform convergence.
For $(c)$ you can just calculate the derivative:
$$f_n(x) = n^\alpha x^n(1-x^2) = n^\alpha(x^n - x^{n+2})$$
$$0 = f_n'(x)= n^\alpha(nx^{n-1} - (n+2)x^{n+1}) \implies x^2 = \frac{n}{n+2}$$
So
$$\sup_{x\in[-1,1]}f_n(x) = f_n\left(\sqrt{\frac{n}{n+2}}\right) = 2\frac{n^{\alpha+\frac{n}2}}{(n+2)^{1+\frac{n}2}}$$
We have
$$2n^{\alpha-1} \le 2\frac{n^{\alpha+\frac{n}2}}{(n+2)^{1+\frac{n}2}} \le 2 (n+2)^{\alpha-1}$$
so $\sup_{n\in\mathbb{N}} \sup_{x\in[-1,1]}f_n(x)$ is finite if and only if $\alpha - 1 < 1$ i.e. $\alpha < 0$.
A: You don't need to get rid of $n^{\alpha},$ you need to bound the term $n^\alpha |x|^n$ uniformly in $x.$ Note it is less than $n^{\alpha}|1-\delta|^n.$ Then to apply the M test, you just need to show $\sum_n n^{\alpha}|1-\delta|^n$ converges.
