Finding the radius of convergence

In summary, the conversation is about finding the radius of convergence for the given series using the ratio test. The correct expression is ((2x)^n)/n!, and the ratio test gives a limit of zero, indicating that the series converges for any value of x. However, one of the participants realizes they were working on the wrong problem and corrects it to be 1+2x+((4!x^(2))/(2!))+((6!x^(3))/((3!^(2))+..., which changes the answer to (1/4). They then ask for help finding the correct series, which is solved by realizing that the numerator is a sequence of even factorials.
  • #1
Timebomb3750
59
0
Finding the radius of convergence...

Homework Statement



1+2x+(4x^(2)/2!)+(8x^(3)/3!)+(16x^(4)/4!)+(32x^(5)/5!)+...

Homework Equations



I would use the ratio test. Which is...

lim as n→∞ (An+1/An)

The Attempt at a Solution



I know what to do to find the answer, but I don't know how to find the expression for the provided series.

I'm guessing it's ((2x^(n))/n!), then work with the ratio test from there?
 
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  • #2


First, I assume you mean ((2x)^n)/n!. That IS correct. Use the ratio test from here. Remember that the ratio has to be less than 1 (as n goes to infinity) for the series to converge. That should help you out.
 
  • #3


Char. Limit said:
First, I assume you mean ((2x)^n)/n!. That IS correct. Use the ratio test from here. Remember that the ratio has to be less than 1 (as n goes to infinity) for the series to converge. That should help you out.

Sweet. I was right on the expression. Okay, so I did the ratio test.

(2x^(n+1))/((n+1)!) * (n!/2x^(n)) = ((xn!)/(n+1)!) But doesn't this lim as n→∞ of this equal 0? Or is there more simplification I need to do? The answer is (1/4), but I don't see how I'm going to get that.
 
  • #4


First, the ratio test gives zero, i.e. less than 1 for any x you choose, so what does that tell you about the radius?

1/4 is definitely not the correct answer for this function. I checked that two different ways, and they gave me the same answer... and it wasn't 1/4.
 
  • #5


Char. Limit said:
First, the ratio test gives zero, i.e. less than 1 for any x you choose, so what does that tell you about the radius?

1/4 is definitely not the correct answer for this function. I checked that two different ways, and they gave me the same answer... and it wasn't 1/4.

Odd. Because the textbook says the answer to this is (1/4). Well. I should double check to see if my expression agrees with the given series.
 
  • #6


Oops! Wow. I was doing the wrong problem.

The series is 1+2x+((4!x^(2))/(2!))+((6!x^(3))/((3!^(2))+...

Sorry. Okay. I know the denominator of the expression is going to be (n!)^2, but not so sure of the numerator. Other than that, I probably can handle the rest of the problem.
 
  • #7


Timebomb3750 said:
Oops! Wow. I was doing the wrong problem.

The series is 1+2x+((4!x^(2))/(2!))+((6!x^(3))/((3!^(2))+...

Sorry. Okay. I know the denominator of the expression is going to be (n!)^2, but not so sure of the numerator. Other than that, I probably can handle the rest of the problem.

How do I edit posts? I keep typing the series wrong...
 
  • #8


Forget it. Delete this. I'll just make a new thread asking for help on finding the series. I'll be sure to type out the series correctly this time. Sorry. I feel so stupid.
 
  • #9


No need to worry I figured out what you meant. The denominator is (n!)^2, like you said. The numerator isn't that hard, either, actually. Just think about what sequence goes 0!, 2!, 4!, 6!.
 

1. What is the radius of convergence?

The radius of convergence is a mathematical concept used in power series. It represents the distance from the center of the series to the point at which the series converges. In other words, it is the maximum distance from the center at which the series will still converge.

2. How is the radius of convergence calculated?

The radius of convergence is typically calculated using the ratio test or the root test. These tests involve taking the limit of the ratio or root of successive terms in the series. If the limit is less than 1, the series will converge, and the radius of convergence can be determined from the limiting value.

3. What is the significance of the radius of convergence?

The radius of convergence is significant because it determines the values of x for which the power series will converge. It also provides information about the behavior of the series as x approaches the radius, such as whether the series will converge or diverge at that point.

4. Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. It is always a positive value, representing the distance from the center of the series to the point of convergence. If the series does not converge at any points, the radius of convergence is said to be 0.

5. How does the radius of convergence relate to the interval of convergence?

The interval of convergence is the set of all x-values for which the power series converges. The radius of convergence is the distance from the center to the endpoints of this interval. In other words, the interval of convergence will always be centered at the point where the series converges, and the radius of convergence will determine the width of this interval.

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