# Homework Help: Using ratio test to find radius of convergence

1. Dec 7, 2011

### Timebomb3750

1. The problem statement, all variables and given/known data
Ʃ((x-3)^(n)) / (n*2^(n))

2. Relevant equations

lim as n→ ∞ (An+1 / An)

3. The attempt at a solution

When dividing two fractions, invert the second and multiple to get what you see below.

(x-3)^(n+1)/((n+1)*2^(n+1)) * (n*2^(n))/((x-3)^(n))

Do some cross canceling to get (n(x-3)) / (2(n-1))

Then when you take the limit as n goes to infinity, you're just left with (x-3)/2

What I'm stuck on is what to do next. The answer is two, but I'm not seeing it.

2. Dec 7, 2011

### genericusrnme

You use the fact that the absolute value of the ratio must be less than or equal to 1 to find x

$\left|\frac{1}{2} (-3+x)\right|<1$

$| (-3+x)|< 2$

the radius of convergence is 2, and it is centered at x=3

3. Dec 7, 2011

### micromass

First of all, you forgot absolute value signs.

Secondly, what does the ratio test state??

4. Dec 7, 2011

### Timebomb3750

Oh okay. I thought you had to get x by itself, so the radius of convergence would be 5.

5. Dec 7, 2011

### micromass

No, it is not 5. Read what he wrote again.

6. Dec 7, 2011

When you get x by itself you would be solving the inequality and thus finding the interval of convergence which is different from the radius of convergence.

7. Dec 7, 2011

### Timebomb3750

I know it's not 5. That's what I assumed it was before I posted this thread. But I don't understand what the role of the -3 is.

8. Dec 7, 2011

To make the series centered at 3.

9. Dec 7, 2011

### HallsofIvy

It is the center point of the interval of convergence. If the radius of convergence is 2, then the series converges between 3- 2= 1 and 3+ 2= 5.

10. Dec 7, 2011

### genericusrnme

the -3 is just the same as a graph of y = (x-3)
in that case it centers the line at x=3

in your case it centers the circle of convergence at x=3

11. Dec 7, 2011

### Timebomb3750

Thanks for clearing that up. But the problem only asked for radius on convergence.