Find the Interval of Convergence

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Homework Help Overview

The discussion revolves around finding the interval of convergence for the series \(\sum^{∞}_{n=0} \frac{(x-5)^n}{n^4 * 2^n}\). Participants are exploring the application of the ratio test and its implications for convergence.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the ratio test and questions the outcome when the limit approaches zero. Some participants clarify the limit's value and its implications for the radius of convergence. Others express confusion regarding the distinction between the radius and interval of convergence.

Discussion Status

Participants are actively engaging with the problem, providing clarifications and corrections regarding the calculations. There is a focus on understanding the relationship between the radius and interval of convergence, with multiple interpretations being explored.

Contextual Notes

Some participants mention the importance of including factors in their calculations, and there is an ongoing discussion about the correct interval of convergence based on the derived limits.

whatlifeforme
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Homework Statement


Find the interval of convergence.

Homework Equations


[itex]\displaystyle \sum^{∞}_{n=0} \frac{(x-5)^n}{n^4 * 2^n}[/itex]

The Attempt at a Solution


I used the ratio test as follows:

[itex]\displaystyle \frac{(x-5)^{n+1}}{(n+1)^4 * 2^{n+1}} * \frac{n^4 2^n}{(x-5)^n}[/itex]

taking the limit:

(x-5) lim (x->inf) [itex]\displaystyle \frac{n^4}{2(n+1)^4}[/itex] = 0

Most of these problems the limit comes to 1. How do i solve this now that the limit is zero?
 
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You are very much on the right track, except that
[tex]\lim_{n \to \infty} \frac{n^4}{(n + 1)^4} \neq 0[/tex]

You can see this if you expand [itex](n + 1)^4 = n^4 + \mathcal O(n^3)[/itex]

(Don't forget the factor of 1/2 in the end)
 
|x-5| < 1

-1 < x-5 < 1
4 < x < 6

Answer (interval of convergence): 4 < x < 6

Does that look right?
 
i think i forgot the factor of (1/2), right?
 
Yes, you did. The correct limit is 1/2. Now, what does that tell you about the radius of convergence?
 
HallsofIvy said:
Yes, you did. The correct limit is 1/2. Now, what does that tell you about the radius of convergence?

what is the radius of convergence vs interval of convergence?
 
whatlifeforme said:
|x-5| < 1

-1 < x-5 < 1
4 < x < 6

Answer (interval of convergence): 4 < x < 6

Does that look right?

is this part correct? because that is one of the choices as the answer but 2 < x < 3 is not.
 
bump.
 
whatlifeforme said:
what is the radius of convergence vs interval of convergence?

The radius of convergence is the "width" of the interval of convergence. E.g. if you are looking at the series around x = 2, if the radius of convergence is 4 the interval of convergence would be 2 - 4 < x < 2 + 4, i.e. -2 < x < 6.

HallsofIvy said:
Yes, you did. The correct limit is 1/2. Now, what does that tell you about the radius of convergence?
First try answering Halls' question. The radius of convergence is related to the ratio (1/2) you calculated.

whatlifeforme said:
bump.
Please don't do that, we are not online 24 hours per day and we will get back to you when we have time.
 
  • #10
sorry for the bump. I'm getting 3 < x < 7 now that I added in the factor of 1/2

-1 < (x-5)/2 < 1
 

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