# Hint requested for: If $\sum_{n=0}^{\infty} a_n x^n$ converges for some $x_0$, then it converges uniformly and absolutely on $[-a, a]$ with $a<|x_0|$?

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 $$0. (Sorry not enough characters to put $$0 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.

• What about $\sum_{n=1}^\infty x^n/n$ and $x_0=-1$? – Angina Seng Dec 28 '18 at 16:00
• OP should say if "$\sum a_nx^n$ converges absolutely for some $x_0$." – D_S Dec 28 '18 at 16:42

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 $$a<\lvert x_0\rvert$$.
• 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. – Ovi Dec 28 '18 at 16:27
• Of course we can't! Pick $M_n=\lvert a_n\rvert a^n$ instead. – José Carlos Santos Dec 28 '18 at 16:29
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 $$n\to \infty.$$