Series-Interval of Convergence

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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?
 
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.
 
What is eqn?

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