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steelphantom Apr29-08 10:08 AM

Real Analysis - Radius of Convergence
 
1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.

Dick Apr29-08 10:14 AM

Right. You can't use the alternating series test. How about a comparison test?

steelphantom Apr29-08 10:30 AM

Thanks! So since [tex]\sum[/tex]anRn converges, and an(-R)n <= anRn for all n, then [tex]\sum[/tex]an(-R)n converges.

Dick Apr29-08 10:32 AM

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?

steelphantom Apr29-08 11:19 AM

Quote:

Quote by Dick (Post 1710016)
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?

Dick Apr29-08 11:27 AM

Nothing wrong with that. Unfortunately, I wasn't thinking of that comparison test. Hence the panic attack. Carry on.

steelphantom Apr29-08 11:28 AM

Ok! :smile: Thanks once again for your help.

HallsofIvy Apr29-08 12:15 PM

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?

steelphantom Apr29-08 12:37 PM

Quote:

Quote by HallsofIvy (Post 1710133)
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?

Dick Apr29-08 01:30 PM

Quote:

Quote by HallsofIvy (Post 1710133)
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.


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