Using ratio test to find radius of convergence

In summary: I don't see how the -3 would help. Thanks for clearing that up. But the problem only asked for radius on convergence. I don't see how the -3 would help.
  • #1
Timebomb3750
59
0

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.
 
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  • #2
You use the fact that the absolute value of the ratio must be less than or equal to 1 to find x

[itex]\left|\frac{1}{2} (-3+x)\right|<1[/itex]

[itex]| (-3+x)|< 2[/itex]

the radius of convergence is 2, and it is centered at x=3
 
  • #3
Timebomb3750 said:

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??
 
  • #4
genericusrnme said:
You use the fact that the absolute value of the ratio must be less than or equal to 1 to find x

[itex]\left|\frac{1}{2} (-3+x)\right|<1[/itex]

[itex]| (-3+x)|< 2[/itex]

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.
 
  • #5
Timebomb3750 said:
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.
 
  • #6
Timebomb3750 said:
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.
 
  • #7
micromass said:
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.
 
  • #8
Timebomb3750 said:
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.
 
  • #9
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
Timebomb3750 said:
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
 
  • #11
genericusrnme said:
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.
 

Related to Using ratio test to find radius of convergence

1. What is the ratio test method for finding the radius of convergence?

The ratio test is a method used to determine the convergence or divergence of a infinite series. It involves taking the limit of the absolute value of the ratio of consecutive terms in the series. If the limit is less than 1, the series is convergent and the radius of convergence can be found using this method.

2. How do I know when to use the ratio test to find the radius of convergence?

The ratio test is typically used for series that involve factorials or exponential terms. It is also a good method to use when the terms in the series are not easily recognizable as a specific type of series, such as geometric or telescoping series.

3. Can the ratio test be used for all series?

No, the ratio test is not always applicable and may not give conclusive results for certain series. Other methods, such as the root test or comparison test, may be more suitable for some series.

4. How do I calculate the radius of convergence using the ratio test?

To find the radius of convergence using the ratio test, you must first apply the test and determine if the limit is less than 1. If it is, then the radius of convergence can be found by taking the reciprocal of the limit. This will give you the radius of the largest circle within which the series will converge.

5. What are the limitations of using the ratio test to find the radius of convergence?

The ratio test is a useful tool, but it may not always give accurate results. In some cases, the limit may be equal to 1, making it inconclusive. Additionally, the ratio test may not work for series with alternating signs or for series that do not approach infinity. In these cases, other methods may need to be used to find the radius of convergence.

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