Solving Limits: Find Interval & Radius of Convergence

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Homework Help Overview

The discussion revolves around finding the radius of convergence and the interval of convergence for a series. Participants are exploring the conditions under which the series converges, with an initial assertion that the series converges for every real number, leading to an interval of (-∞, ∞) and a radius of convergence of ∞.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to take the limit of the entire term in the summation, including factors like (x + 1)n. There are questions about how to demonstrate that the limit equals 0 and how to simplify factorial expressions. Some participants express uncertainty about the rules and calculations involved.

Discussion Status

Guidance has been offered regarding the simplification of factorials and the importance of including all relevant terms in limit calculations. Participants are actively engaging with each other's suggestions and clarifying their understanding of the problem, though no consensus has been reached on the final approach.

Contextual Notes

There is mention of homework constraints, including the need to avoid posting work as images and the distinction that this problem is not considered precalculus. Some participants express confusion about the definitions and rules applicable to the problem.

ironman
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Homework Statement


[/B]
I have to find the radius of convergence and convergence interval. So for what x's the series converge.
The answer is supposed to be for every real number. So the interval is: (-∞, ∞).
So that must mean that the limit L = 0. So the radius of convergence [ which is given by R = 1/L ] is going to be ∞.

Homework Equations


[/B]
CodeCogsEqn-2.gif
CodeCogsEqn.gif
(?)

The Attempt at a Solution


[/B]
I tried using
CodeCogsEqn-4.gif
with an =
CodeCogsEqn-3.gif


but i don't seem to get anywhere...
Is there another way of showing that the lim = 0 ?
 
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ironman said:

Homework Statement


[/B]
I have to find the radius of convergence and convergence interval. So for what x's the series converge.
The answer is supposed to be for every real number. So the interval is: (-∞, ∞).
So that must mean that the limit L = 0. So the radius of convergence [ which is given by R = 1/L ] is going to be ∞.

Homework Equations


[/B]
View attachment 76309View attachment 76306 (?)
You need to take the limit of the entire term in the summation, including the (x + 1)n factor.
ironman said:

The Attempt at a Solution


[/B]
I tried using View attachment 76307 with an = View attachment 76308

but i don't seem to get anywhere...
Is there another way of showing that the lim = 0 ?
Show us the work that you did in taking the limit of an+1/an. It's easy to get wrong with those factorials.

BTW, this is NOT a precalculus problem, so I moved it to the calculus section.
 
Mark44 said:
You need to take the limit of the entire term in the summation, including the (x + 1)n factor.

Show us the work that you did in taking the limit of an+1/an. It's easy to get wrong with those factorials.

BTW, this is NOT a precalculus problem, so I moved it to the calculus section.

Ah ok I'm sorry.
Yes indeed. But you would get:
CodeCogsEqn-5.gif

So you might as well just use the 'n-part', for the (x-c) is just a number (?)

I ended up with:

CodeCogsEqn-6.gif
It's not possible to simplify this, right?
 
ironman said:
Ah ok I'm sorry.
Yes indeed. But you would get: View attachment 76323
So you might as well just use the 'n-part', for the (x-c) is just a number (?)

I ended up with:

View attachment 76324 It's not possible to simplify this, right?

Sure you can simplify it. Start with the (n+1)!/n! part. Remember what the definition of n! and (n+1)! are. A lot of terms cancel.
 
Please don't post your work as images. When I quote what you wrote your stuff doesn't show up.
ironman said:
Ah ok I'm sorry.
Yes indeed. But you would get: View attachment 76323
Actually you get |x + 1|. It needs to be included in your limit calculations, especially for when you are finding the interval of convergence.
.
ironman said:
So you might as well just use the 'n-part', for the (x-c) is just a number (?)

I ended up with:

View attachment 76324 It's not possible to simplify this, right?
It can and should be simplified. (n + 1)! = (n + 1) * n! and (2n + 2)! = (2n + 2)(2n + 1)*(2n)!.
 
Dick said:
Sure you can simplify it. Start with the (n+1)!/n! part. Remember what the definition of n! and (n+1)! are. A lot of terms cancel.

Ah of course!
(n+1)!/ n! = n+1.

I still don't know these rules. But I used :
n=3
4*3*2*1/3*2*1 = 4
= n+1
 
Mark44 said:
Please don't post your work as images. When I quote what you wrote your stuff doesn't show up.Actually you get |x + 1|. It needs to be included in your limit calculations, especially for when you are finding the interval of convergence.
.
It can and should be simplified. (n + 1)! = (n + 1) * n! and (2n + 2)! = (2n + 2)(2n + 1)*(2n)!.

Mark44 said:
Please don't post your work as images. When I quote what you wrote your stuff doesn't show up.Actually you get |x + 1|. It needs to be included in your limit calculations, especially for when you are finding the interval of convergence.
.
It can and should be simplified. (n + 1)! = (n + 1) * n! and (2n + 2)! = (2n + 2)(2n + 1)*(2n)!.

Hmm I see.
You would get (n+1) * (2n+1) * (2n+2) * (4n)! / (4n +4)!
Can you divide both the nom. and denom. by (4n+4)! and end up with 0/1 = 0 as [ n -> ∞ ] ?

Weird that my calculus book leaves the |x +c| out then...
 
ironman said:
Hmm I see.
You would get (n+1) * (2n+1) * (2n+2) * (4n)! / (4n +4)!
Can you divide both the nom. and denom. by (4n+4)! and end up with 0/1 = 0 as [ n -> ∞ ] ?

Weird that my calculus book leaves the |x +c| out then...

You should keep simplifying until you have no more factorials. Take care of the (4n)!/(4n+4)! part as well.
 
Dick said:
You should keep simplifying until you have no more factorials. Take care of the (4n)!/(4n+4)! part as well.

I've no idea how to do that...
 
  • #10
ironman said:
I've no idea how to do that...

Isn't it the case that ##(4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1) (4n)!##? Just keep using the definition of factorial.
 
  • #11
Ray Vickson said:
Isn't it the case that ##(4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1) (4n)!##? Just keep using the definition of factorial.

Oh well... I blame it on my lack of sleep, ha.
So I end up with polynomials only. 4-degree in the denominator and 3-degree in the numerator. (highest degrees)
 
  • #12
ironman said:
Oh well... I blame it on my lack of sleep, ha.
So I end up with polynomials only. 4-degree in the denominator and 3-degree in the numerator. (highest degrees)

Ok, so do you know how to find its limit?
 
  • #13
Dick said:
Ok, so do you know how to find its limit?

Yes, I do. Thanks for the help all!
 

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