Series-Interval of Convergence

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Homework Help Overview

The discussion revolves around determining the radius of convergence for the series \(\Sigma_{n=0}^{\infty} \frac{(x-4)^{2n}}{(n+1)(11^n)}\). Participants are exploring the application of convergence tests and the implications of their findings.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the ratio test to find the radius of convergence, with some attempting to rewrite the series in terms of a new variable. Questions arise about the interpretation of limits and the conditions for convergence.

Discussion Status

There is active exploration of different interpretations of the series and its convergence criteria. Some participants have offered partial insights into the limits derived from their calculations, while others express confusion about the implications of their results and the nature of the bounds for \(x\).

Contextual Notes

Participants are grappling with the constraints of the problem, including the nature of the series and the assumptions about the values of \(x\). There is uncertainty regarding the validity of certain steps and the interpretation of negative values in the context of the convergence conditions.

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Homework Statement


[itex]\Sigma[/itex] from n=0 to infinity (x-4)^(2n)/((n+1)(11^n))
Find the Radius of Convergence


Homework Equations



limit of (Cn/(Cn+1)
Which I found it to be 11 (isn't this supposed to be the Radius of Convergence?)


The Attempt at a Solution

 
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To find the radius of convergence of a geometric series [itex]\sum_{n=0}^{\infty}a_n[/itex] you take the following limit and solve for x, or an expression involving x.
[tex]a_{n}=\frac{(x-4)^{2n}}{(n+1)11^{n}}[/tex]
[tex]\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1[/tex]
[tex]\lim_{n\rightarrow\infty}\left|\frac{(x-4)^{2(n+1)}}{(n+2)11^{n+1}}\cdot\frac{(n+1)11^{n}}{(x-4)^{2n}}\right|<1[/tex]
Can you continue from here?

Edit: I inserted the absolute value symbols that I originally forgot.
 
Last edited:
1LastTry said:

Homework Statement


[itex]\Sigma[/itex] from n=0 to infinity (x-4)^(2n)/((n+1)(11^n))
Find the Radius of Convergence


Homework Equations



limit of (Cn/(Cn+1)
Which I found it to be 11 (isn't this supposed to be the Radius of Convergence?)


The Attempt at a Solution


Let y = (x-4)^2, and re-write the series.
 
From what u guys said I got ((x-4)^2)/11
 
1LastTry said:
From what u guys said I got ((x-4)^2)/11

Ok, so how does that translate into radius of convergence? Where is |(x-4)^2/11|<1?
 
well i got sqr(-11) +4 < x < 15?
 
1LastTry said:
well i got sqr(-11) +4 < x < 15?

That's pretty wrong. Can you show how you got it??!
 
well

-1< (x-4)^2/11 < 1 (is this right?)

then i solved that and got to that soltuion. I had a feeling that its wrong since it came out with some weir dnumbers

could it be -7<x<15
 
1LastTry said:
well

-1< (x-4)^2/11 < 1 (is this right?)
Is there any way how (x-4)^2/11 could be negative? If not, what is the lower limit?

Note that the upper limit (<1) gives two separate limits for x.

could it be -7<x<15
What happens if you set x=14 in the equation above?
 
  • #10
O ok, ummm i stil hve to think about this butI guess it can't be negative, sorry I wasnt thinking. I solved it again and would it be 0< eqn < sqrt(11) +4?

Im still kinda confused on how this works, sorry.
 
  • #11
What is eqn?

Please show all steps you did, otherwise it is hard to guess where you did something wrong.
 

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