Real Analysis - Radius of Convergence

In summary, the conversation discusses how to show that if a series has a finite radius of convergence and all terms are positive, then if it converges at R, it also converges at -R. The discussion explores using the comparison test, the Dirichlet convergence test, and the properties of alternating series to prove this statement. It is ultimately determined that the series is absolutely convergent and therefore the argument using the comparison test is valid.
  • #1
steelphantom
159
0

Homework Statement


Suppose that [tex]\sum[/tex]anxn has finite radius of convergence R and that an >= 0 for all n. Show that if the series converges at R, then it also converges at -R.

Homework Equations



The Attempt at a Solution


Since the series converges at R, then I know that [tex]\sum[/tex]anRn = M.

At -R, the series is the following: [tex]\sum[/tex]an(-R)n = [tex]\sum[/tex](-1)nanRn.

I'm not sure where to go from here. I thought I needed to use the alternating series test, but how can I know that a1 >= a2 >= ... >= an for all n? Do I know this because the series converges? Thanks for your help.
 
Physics news on Phys.org
  • #2
Right. You can't use the alternating series test. How about a comparison test?
 
  • #3
Thanks! So since [tex]\sum[/tex]anRn converges, and an(-R)n <= anRn for all n, then [tex]\sum[/tex]an(-R)n converges.
 
  • #4
Sorry! That's wrong. I'm clearly asleep at the wheel. That's convergence for sequences. And this sort of argument only shows that the partial sums are bounded, not that they converge. Do you know the Dirichlet convergence test?
 
Last edited:
  • #5
Dick said:
Sorry! That's wrong. I'm clearly asleep at the wheel. That's convergence for sequences. And this sort of argument only shows that the partial sums are bounded, not that they converge. Do you know the Dirichlet convergence test?

I don't know that one. But the comparison test in my book says the following:

Let [tex]\sum[/tex]an be a series where an >=0 for all n.
(i) If [tex]\sum[/tex]an converges and |bn| <= an for all n, then [tex]\sum[/tex]bn converges.

If I let an = anRn, this is >=0 for all n. And if I let bn = an(-R)n, then I have |bn| <= an for all n, so the series converges, right? What's wrong with this statement?
 
  • #6
Nothing wrong with that. Unfortunately, I wasn't thinking of that comparison test. Hence the panic attack. Carry on.
 
  • #7
Ok! :smile: Thanks once again for your help.
 
  • #8
What's wrong with using the comparison test is that it only applies to positive series. Certainly -n< (1/2)n for all n, but we can't use that to conclude that [itex]\sum -n [/itex] converges!

The crucial point is that every an is positive. That means that [itex]\sum a_n x^n[/itex], for x negative is an alternating series. What is true of alternating series?
 
Last edited by a moderator:
  • #9
HallsofIvy said:
The crucial point is that every an is positive. That means that [itex]\sum a_n x^n[/itex], for x negative is an alternating series. What is true of alternating series?

If a1 >= a2 >= ... >= an >= ... >= 0 and lim(an) = 0, then the alternating series [tex]\sum[/tex](-1)nan converges. But, like I said before, do I know that a1 >= a2 >= ... >= an because the series converges at R?
 
  • #10
HallsofIvy said:
What's wrong with using the comparison test is that it only applies to positive series. Certainly -n< (1/2)n for all n, but we can't use that to conclude that [itex]\sum -n [/itex] converges!

The crucial point is that every an is positive. That means that [itex]\sum a_n x^n[/itex], for x negative is an alternating series. What is true of alternating series?

All of terms a_n*R^n are positive and it's convergent. The series is absolutely convergent. Nothing can go wrong here.
 

1. What is the definition of radius of convergence in real analysis?

The radius of convergence is a concept in real analysis that refers to the distance from the center of a power series to the nearest point where the series converges. It represents the range of values for which the power series is a valid representation of the function.

2. How is the radius of convergence calculated?

The radius of convergence can be calculated using the ratio test, which involves taking the limit of the absolute value of the ratio of consecutive terms in the power series. If this limit is less than 1, the series converges, and the value of the radius of convergence is the reciprocal of this limit.

3. What is the significance of the radius of convergence in real analysis?

The radius of convergence is an important concept in real analysis because it determines the domain of convergence for a power series. This means that it tells us which values of the independent variable will result in a convergent series, and therefore, a valid representation of the function.

4. Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. It represents a distance and therefore must be a positive value. However, it is possible for the radius of convergence to be infinite, which means that the power series converges for all values of the independent variable.

5. How does the radius of convergence relate to the continuity of a function?

The radius of convergence is closely related to the continuity of a function. If a power series has a positive radius of convergence, the resulting function will be continuous within that radius. This means that the function can be evaluated at any point within the radius of convergence without encountering any discontinuities.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
91
  • Calculus and Beyond Homework Help
Replies
7
Views
653
  • Calculus and Beyond Homework Help
Replies
2
Views
658
  • Calculus and Beyond Homework Help
Replies
1
Views
90
  • Calculus and Beyond Homework Help
Replies
5
Views
940
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
850
  • Calculus and Beyond Homework Help
Replies
6
Views
275
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
753
Back
Top