# Series-Interval of Convergence

1. Jul 8, 2013

### 1LastTry

1. The problem statement, all variables and given/known data
$\Sigma$ from n=0 to infinity (x-4)^(2n)/((n+1)(11^n))

2. Relevant equations

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

3. The attempt at a solution

2. Jul 8, 2013

### Goa'uld

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

Edit: I inserted the absolute value symbols that I originally forgot.

Last edited: Jul 8, 2013
3. Jul 8, 2013

### Ray Vickson

Let y = (x-4)^2, and re-write the series.

4. Jul 8, 2013

### 1LastTry

From what u guys said I got ((x-4)^2)/11

5. Jul 9, 2013

### Dick

Ok, so how does that translate into radius of convergence? Where is |(x-4)^2/11|<1?

6. Jul 9, 2013

### 1LastTry

well i got sqr(-11) +4 < x < 15?

7. Jul 9, 2013

### Dick

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

8. Jul 9, 2013

### 1LastTry

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. Jul 9, 2013

### Staff: Mentor

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.

What happens if you set x=14 in the equation above?

10. Jul 9, 2013

### 1LastTry

I guess it cant 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. Jul 9, 2013

### Staff: Mentor

What is eqn?

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