Uniform convergence for $\sum_{n=1}^\infty n^\alpha x^{2n} (1-x)^2$

Question

$$\sum_{n=1}^\infty n^\alpha x^{2n} (1-x)^2$$

a) Determine for every $\alpha$ where the series converges.
b) Show for every $\alpha$ if there is uniform convergence at $[-1,0]$ or $[0,1]$.

Can anybody help me out here a little bit ? How should I tackle this problem ?

Answer 1

Hint: Use ratio test to see for what values of $\alpha$ the series converges.

Added: For part $(b)$, you need to use the M-test for uniform convergence. Note that

$$ |n^{\alpha} x^{2n}| \leq n^{\alpha},\,\,\, |x|\leq 1 . $$

Now, you should be able to see for what $\alpha$ the series converges uniformly.

Answer 2

Ok, Use Root Test.

$$ \lim_{n\rightarrow \infty} \sqrt[n]{n^{\alpha}x^{2n}(1-x)^2} $$

So, you get both answers from this. Does not depend on $\alpha$ but on $x$.

edit the absolute sign is missing.