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I would like to prove

If $\displaystyle\sum_{n=0}^{\infty} a_n x^n$ converges for some $x_0$, then it converges uniformly and absolutely on $[-a, a]$ with $0<a<|x_0|$. (Sorry not enough characters to put $0<a$ in the title)

I can easily look at the proof in my book, but I really want to learn how to think about analysis and be able to prove these things on my own.

I tried to make it easier by assuming that $x_0>0$ and $\forall (n \in \mathbb{N}) a_n>0$. In this case, we can apply the $M$-Test with $M_n = a_n x_0^n$.

However, I'm not sure how to modify this to work in the general case.

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  • $\begingroup$ What about $\sum_{n=1}^\infty x^n/n$ and $x_0=-1$? $\endgroup$ – Lord Shark the Unknown Dec 28 '18 at 16:00
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    $\begingroup$ OP should say if "$\sum a_nx^n$ converges absolutely for some $x_0$." $\endgroup$ – D_S Dec 28 '18 at 16:42
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Since the series $\displaystyle\sum_{n=0}^\infty a_n{x_0}^n$ converges, you know that $\lim_{n\to\infty}a_n{x_0}^n=0$. This is equivalent to the ssertion that $\lim_{n\to\infty}\lvert a_n{x_0}^n\rvert=0$. This is anough to be able to apply the Weierstrass $M$-test to the series $\displaystyle\sum_{n=0}^\infty a_nx^n$ in $[-a,a]$. Don't forget to use the fact that $a<\lvert x_0\rvert$.

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  • $\begingroup$ But we cannot just pick $M_n = |a_n x_0^n|$, right? Because I think the sum $\sum_{n=0}^\infty |a_n{x_0}^n|$ may not converge. $\endgroup$ – Ovi Dec 28 '18 at 16:27
  • $\begingroup$ Of course we can't! Pick $M_n=\lvert a_n\rvert a^n$ instead. $\endgroup$ – José Carlos Santos Dec 28 '18 at 16:29
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Hint: $|a_nx^n|=\left | a_n\left ( \frac{x}{x_0} \right )^{n}\cdot x^{n}_0 \right |=\left | a_nx_0^{n}\cdot\left ( \frac{x}{x_0} \right )^{n} \right |,\ |x|\le a<|x_0|$ and $\sum a_nx_0^{n} $ converges so $|a_nx_0^{n}|\to 0$ as $n\to \infty.$

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