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

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

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 ?

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I suspect it should be $(1-x)^{2n}$. Otherwise, just get rid of $(1-x)^2$. – 1015 Mar 5 '13 at 17:53
@julien It is $(1-x)^2$. – Kasper Mar 5 '13 at 18:25
Then the factor $(1-x)^2$ has nothing to do with the convergence of the series. You can factor it out. Except for the case $x=1$ which forces the series to converge since the general term becomes $0$. – 1015 Mar 5 '13 at 18:38
@julien For $[-1,0]$ i think there is only convergence at $\alpha<-1$. For $[0,1]$ there must be uniform convergence at $\alpha<-1$, but I'm not sure for $\alpha\geq -1$. – Kasper Mar 5 '13 at 18:47
Like I said, forget $(1-x)^2$ and consider the convergence of $\sum n^\alpha x^{2n}$. This is an even power series. So what happens on $[0,1]$ happens the same way on $[-1,0]$. How can you find differences between these to cases? – 1015 Mar 6 '13 at 0:26

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.

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Shouldn't I take the n'th root. I get that it is convergent at $|x^2|<1$, and divergent at $|x^2|>1$. Convergent at $x=1$ and only convergent at $x=-1$ if $\alpha\leq -1$. Correct ? – Kasper Mar 5 '13 at 18:32
You want to add some absolute values. – 1015 Mar 5 '13 at 18:39
Yes, sorry, it's nth root. Your solution looks good to me. But listen to julien. – user45099 Mar 5 '13 at 18:39

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.

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I've solved the first question, but I don't know how to show uniform convergence at $[0,1]$. Any ideas ? – Kasper Mar 5 '13 at 18:43