Series-Interval of Convergence

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In summary: 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-What happens if you set x=14 in the equation above?-Ok, ummm i stil hve to think about this but I guess it can't be negative.
  • #1
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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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  • #2
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:
  • #3
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.
 
  • #4
From what u guys said I got ((x-4)^2)/11
 
  • #5
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?
 
  • #6
well i got sqr(-11) +4 < x < 15?
 
  • #7
1LastTry said:
well i got sqr(-11) +4 < x < 15?

That's pretty wrong. Can you show how you got it??!
 
  • #8
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
 
  • #9
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.
 

What is a series-interval of convergence?

A series-interval of convergence is a range of values for which a given infinite series will converge. It is typically expressed as an interval on the real number line.

How is the series-interval of convergence determined?

The series-interval of convergence is determined by using the ratio test or the root test. These tests involve taking the limit of the ratio or root of the terms in the series to determine if the series converges or diverges.

Can a series have more than one interval of convergence?

Yes, a series can have multiple intervals of convergence. This occurs when the series has different convergence behavior for different values within the interval.

What does it mean if a series has a finite interval of convergence?

If a series has a finite interval of convergence, it means that the series will only converge for values within that interval. For values outside of the interval, the series will diverge.

Can a series have an empty interval of convergence?

Yes, a series can have an empty interval of convergence. This means that the series does not converge for any values on the real number line and is said to have a radius of convergence of 0.

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