Radius and Interval of convergence

In summary, the given series can be evaluated using the ratio test, which shows that the radius of convergence is 1 and the interval of convergence is (-1,1). When checking the endpoint x=1, the alternating series test can be used, but it is necessary to prove that 1/(n+2) < 1/(n+1) for all n. One way to do this is to start with the inequality 2>1, add n to both sides, and divide by n+2 to get the desired inequality. Another approach is to consider the function f(n) = 1/(n+1) and show that its derivative is negative, indicating that the function is decreasing and thus bn+1 <
  • #1
Painguy
120
0

Homework Statement


Ʃ(((-1)^n)(x^n))/(n+1) from 0 to ∞

Homework Equations


The Attempt at a Solution


I took applied the ratio test and got that lim|(x^(n+1))/(n+2) * (n+1)/(x^n)| =|x|
so that means for it to converge |x|<1 Radius of convergence is 1
My interval is (-1<x<1)
Now I check the endpoints

x=1
Ʃ(((-1)^n))/(n+1) from 0 to ∞
I want to use the alternating series test, but I have run into a little problem.
I know that the Lim 1/(n+1) =0 but how do I prove that 1/(n+2) < 1/(n+1) for all n.
I know its true, but my professor wants us to show that. I can show that -1<1 which is a true statement, but what does that say about the inequality in relation to n?

Is there a better test to do this? Ratio test just equals 1 so that tells me nothing. I can't really use the integral test for an alternating series since it would end up imaginary. If there is a better test could someone still show how to use the alternating series if possible?
 
Physics news on Phys.org
  • #2
Usually you can assume certain properties of real numbers to be true. If your professor really expects you to prove such a trivial point, why don't you start with something "obviously" true, like n+2 > n+1 for all n?
 
  • #3
Or, to start from the very obvious 2> 1. Add n to both sides, then divide both sides by ...
 
  • #4
I set 1/(n+1) to a function f which I then siad was less than 0. when i took the derivative of f i obtained a negative function which is less than 0. I stated that since the derivative is less than zero then the function is decreasing and bn+1 < bn. Is that ok?
 

1. What is the radius of convergence?

The radius of convergence is a measure of how far away from the center of a power series the series remains convergent. It is denoted by R and can be calculated using the ratio test or the root test.

2. How do I determine the interval of convergence?

To determine the interval of convergence, you need to find the values of x for which the power series converges. This can be done by using the ratio test or the root test and checking the endpoints of the interval for convergence.

3. Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. It is always a positive number or infinity.

4. What happens if the radius of convergence is infinite?

If the radius of convergence is infinite, it means that the power series converges for all values of x. This is also known as the entire interval of convergence.

5. How is the radius of convergence related to the convergence of a power series?

The radius of convergence is directly related to the convergence of a power series. If the absolute value of x is less than the radius of convergence, then the series will converge. If the absolute value of x is greater than the radius of convergence, then the series will diverge.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
160
  • Calculus and Beyond Homework Help
Replies
1
Views
218
  • Calculus and Beyond Homework Help
Replies
2
Views
696
  • Calculus and Beyond Homework Help
Replies
4
Views
264
  • Calculus and Beyond Homework Help
Replies
5
Views
970
  • Calculus and Beyond Homework Help
Replies
2
Views
760
  • Calculus and Beyond Homework Help
Replies
2
Views
766
  • Calculus and Beyond Homework Help
Replies
2
Views
939
  • Calculus and Beyond Homework Help
Replies
3
Views
397
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top