Power series, interval of convergence

Question

Let a power series be $$\sum_{n=0}^{\infty}a_nx^n$$ and if $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=0$$, then is it true that the power series converges for all $x \in \mathbb{R}$?

If that limit has the absolute value, then using the Ratio Test, this is indeed true, but does it work without the absolute value?

mathcounterexamples.net · Accepted Answer · 2020-12-09T21:40:47.020

2

If the limit is zero without absolute value, then it is also zero with absolute value.... as the absolute value map is continuous at zero.

So you get that the radius of convergence is infinite and that your power series converges everywhere.

edited Dec 09 '20 at 21:40

answered Dec 09 '20 at 21:26

mathcounterexamples.net

70,018

what if there is a $(-1)^n$ in front of the series? – Logan Dec 09 '20 at 21:28
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The absolute value of $(-1)^n$ is $1$ so there is no issue. Zero multiplied by one or minus one is still zero. – mathcounterexamples.net Dec 09 '20 at 21:30
The thing that worries me is that, I know absolute convergence is stronger than convergence, so if that limit converges absolutely, then it converges, but does that works backwards? – Logan Dec 09 '20 at 21:32
so the answer is true? – Logan Dec 09 '20 at 21:33
@Logan not always. Consider $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$ – Alann Rosas Dec 09 '20 at 21:33
em... so my statement is not true? – Logan Dec 09 '20 at 21:36
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@Logan which statement? The claim in your post, or the claim that ordinary convergence implies absolute convergence? – Alann Rosas Dec 09 '20 at 21:39
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For sure it is true! Your power series converges everywhere as $\lim \vert \frac{a_{n+1}}{a_n} \vert = 0$. – mathcounterexamples.net Dec 09 '20 at 21:39
oh!!! right! that makes sense, my oogonial statement is true, my comment statement is not. – Logan Dec 09 '20 at 21:41
Thank you guys so much! – Logan Dec 09 '20 at 21:41

score 0 · Answer 2 · answered Dec 09 '20 at 21:26

0

In general absolute convergence is stronger than convergence. So knowing that $\sum_{n=0}^\infty |a_n|x^n$ converges for all n, then this tells us that $\sum_{n=0}^\infty a_nx^n$ must also converge. for example by the squeeze theorem.

answered Dec 09 '20 at 21:26

Moosh

624

Power series, interval of convergence

2 Answers2