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Root test convergence

Anonymous Thu 12 Aug 13:27:58 2021 No.13509127 View Reply Original Report

Quoted By: >>13509209 >>13509528 >>13512646

Can somebody explain where the ? came from and why (a_n)^1/n*r<1-? and why does that mean convergence?

Anonymous

Anonymous Thu 12 Aug 2021 13:51:04 No.13509209 Report

Quoted By:

>>13509127
Bump

Anonymous

Anonymous Thu 12 Aug 2021 14:08:44 No.13509264 Report

Quoted By:

Bump

Anonymous

Anonymous Thu 12 Aug 2021 14:22:50 No.13509315 Report

Quoted By:

Can anyone explain this?

Anonymous

Anonymous Thu 12 Aug 2021 14:35:00 No.13509361 Report

Quoted By:

Bump

El Arcón

El Arcón Thu 12 Aug 2021 14:47:43 No.13509404 Report

Quoted By: >>13509410 >>13509484

It comes from the Cauchy definition of a limit. If the series a(n) converges to s, then for any s-epsilon there exists an N such that for all m>N we have
a(m) > s-epsilon.

This is the property stated in the proof. It follows because the limit identity in the theorem statement gives a(n) as a Cauchy sequence. Those kinds of sequences have this property.

El Arcón

El Arcón Thu 12 Aug 2021 14:48:30 No.13509410 Report

Quoted By:

>>13509404
*increasing* sequence

Anonymous

Anonymous Thu 12 Aug 2021 15:05:34 No.13509484 Report

Quoted By: >>13509507

>>13509404
>If the series a(n) converges to s
I don't think it does, s is just what (a_n)^1/n becomes in the limit as n->?.

El Arcón

El Arcón Thu 12 Aug 2021 15:11:06 No.13509507 Report

Quoted By: >>13509514 >>13509522

>>13509484
a(n) = (a_n)^1/n

Anonymous

Anonymous Thu 12 Aug 2021 15:12:29 No.13509514 Report

Quoted By:

>>13509507
Why is a(n)r<1-? instead of a(n)r<1?
For all n sufficiently large doesn't a(n) become s and sr<1?

El Arcón

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El Arcón Thu 12 Aug 2021 15:15:18 No.13509521 Report

Quoted By: >>13509531

This is an example of what I mean when I say that contrary to the way most people are read, people read me not with the intention to understand what I wrote but instead read me with intention to say that what I wrote is wrong.
>limit identity in the theorem statement gives a(n) as a Cauchy sequence.

Anonymous

Anonymous Thu 12 Aug 2021 15:15:40 No.13509522 Report

Quoted By: >>13509581

>>13509507
Also why do they say that the series converges by comparison to the geometric series?

El Arcón

El Arcón Thu 12 Aug 2021 15:17:52 No.13509526 Report

Quoted By:

Do you see what I mean? There's only thing in there given as a Cauchy sequence. Instead of understanding that I have labelled that thing as a(n), the choose the thing not labelled a(n) and then say, "Nuh-uh. That's not a Cauchy sequence."

Anonymous

Anonymous Thu 12 Aug 2021 15:18:05 No.13509528 Report

Quoted By: >>13509548

>>13509127
They just pick any epsilon for which the inequality holds. For example you can choose
$\varepsilon = \frac{1 - rs}{2}$ . Any smaller epsilon will work too.
$a_n^{1/n}$ converges to $s$ so $a_n^{1/n} r$ converges to $sr$
Then by definition of convergence for the previous chosen epsilon there is an $N$ such that for all $n > N$
$\left|a_n^{1/n} r - sr \right| < \varepsilon$
This implies for those n that
$a_n^{1/n} r< sr + \varepsilon = 1 - \varepsilon$

This means convergence since
$\sum_{n=0}^\infty a_n r^n \\ = \sum_{n=0}^{N-1} a_n r^n + \sum_{n=N}^{\infty} a_n r^n \\ < \sum_{n=0}^{N-1} a_n r^n + \sum_{n=N}^{\infty} (1-\varepsilon)^n \\ = \sum_{n=0}^{N-1} a_n r^n + \frac{(1-\varepsilon)^N}{1 - (1-\varepsilon)} \\ = \sum_{n=0}^{N-1} a_n r^n + \frac{(1-\varepsilon)^N}{\varepsilon} \\ < \infty $

Anonymous

Anonymous Thu 12 Aug 2021 15:18:27 No.13509531 Report

Quoted By:

>>13509521
Sorry, I haven't encountered anything called a 'cauchy something' in the entire book (a first course in calculus). So I'm not entirely sure what you mean.

Anonymous

Anonymous Thu 12 Aug 2021 15:20:31 No.13509540 Report

Quoted By:

OP, El Arcon is the resident schizo. Don't take him too seriously.

Anonymous

Anonymous Thu 12 Aug 2021 15:22:34 No.13509548 Report

Quoted By: >>13509551

>>13509528
Thanks, I think this makes sense. But did you mean that ((a_n)^1/n)r<sr+?<1-??
Instead of ((a_n)^1/n)r<sr+?=1-?

Anonymous

Anonymous Thu 12 Aug 2021 15:23:35 No.13509551 Report

Quoted By:

>>13509548
Shit nevermind, I misread it. Sorry

El Arcón

El Arcón Thu 12 Aug 2021 15:32:11 No.13509581 Report

Quoted By: >>13509586 >>13509610

>>13509522
Since the terms
a(n)=a_n^{1/n}

approach s, we know that each term in the compared series is less than the corresponding power term in the geometric series whose coefficients are s. That series converges for r<1 which is given by the condition
r < 1/s

El Arcón

El Arcón Thu 12 Aug 2021 15:33:38 No.13509586 Report

Quoted By:

>>13509581
0< r < 1/s

Anonymous

Anonymous Thu 12 Aug 2021 15:39:05 No.13509606 Report

Quoted By:

This is trivial bro. By convergence, any open ball/interval around rs contains almost all terms of the sequence. So just take the open ball of radius (1-rs)/2. Thus almost all terms are bounded by y^n for y=1-epsilon<1 fixed.

El Arcón

El Arcón Thu 12 Aug 2021 15:39:33 No.13509610 Report

Quoted By:

>>13509581
Actually, the condition that gives the convergence is not that r<1 but that the exponentiated combination of the r and the a(n) are less than one, which is given by the condition r < 1/s

El Arcón

El Arcón Thu 12 Aug 2021 15:41:57 No.13509618 Report

Quoted By: >>13509628

Trivial means "immediately easy" for someone who is well practiced but trivial means "possible to understand" for someone who has spend a couple of hundred hours doing analysis in the last ten years.

Anonymous

Anonymous Thu 12 Aug 2021 15:43:19 No.13509628 Report

Quoted By: >>13509630 >>13509635

>>13509618
Brainlet cope

El Arcón

El Arcón Thu 12 Aug 2021 15:44:37 No.13509630 Report

Quoted By: >>13509657

>>13509628
So you think that immediately seeing a solution is depends on brainpower alone in the absence of practice? If it's trivial, brainpower has literally nothing to do with it.

El Arcón

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El Arcón Thu 12 Aug 2021 15:45:52 No.13509635 Report

Quoted By:

>>13509628
>Brainlet cope

Anonymous

Anonymous Thu 12 Aug 2021 15:52:37 No.13509657 Report

Quoted By:

>>13509630
Brainlet schizo pseud ramblings. Stick to bullshitting in physics

El Arcón

El Arcón Thu 12 Aug 2021 16:40:39 No.13509904 Report

Quoted By:

You ought to believe what is written about the fear of the Lord.

Anonymous

Anonymous Fri 13 Aug 2021 05:57:43 No.13512646 Report

Quoted By:

>>13509127
This is pretty straightforward.
All you are doing is using geometric series convergence to verify convergence of power series for an interval of values.
You are arguing that the magnitude of $a_{n}x^{n}$ is bounded above by $c^{n}$ , where c is a positive, real number smaller than 1. Hence, the absolute power series is bounded above by the geometric series sum. Because the power series absolutely converges on an open interval, it must be convergent (this fact is obvious if you prove the contrapositive on the cauchy sequence of partial sums).
$c = 1 - \epsilon$ .

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