Using ratio test to find radius of convergence

[Timebomb3750]

Homework Statement

Ʃ((x-3)^(n)) / (n*2^(n))

Homework Equations

lim as n→ ∞ (An+1 / An)

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.

 

[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

 

[micromass]

Homework Statement

Ʃ((x-3)^(n)) / (n*2^(n))

Homework Equations

lim as n→ ∞ (An+1 / An)

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.

First of all, you forgot absolute value signs.

Secondly, what does the ratio test state??

 

[Timebomb3750]

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

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

 

[micromass]

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

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

 

[DeadOriginal]

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

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.

 

[Timebomb3750]

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

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.

 

[DeadOriginal]

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.

To make the series centered at 3.

 

[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.

 

[genericusrnme]

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.

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

 

[Timebomb3750]

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

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