Real Analysis - Radius of Convergence

steelphantom

1. The problem statement, all variables and given/known data
Suppose that [tex]\sum[/tex]a_nxⁿ has finite radius of convergence R and that a_n >= 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]a_nRⁿ = M.

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

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

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

steelphantom

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

Dick

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

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

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

If I let a_n = a_nRⁿ, this is >=0 for all n. And if I let b_n = a_n(-R)ⁿ, then I have |b_n| <= a_n for all n, so the series converges, right? What's wrong with this statement?

Dick

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

steelphantom

Ok! Thanks once again for your help.

HallsofIvy

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 a_n 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

If a1 >= a2 >= ... >= an >= ... >= 0 and lim(an) = 0, then the alternating series [tex]\sum[/tex](-1)ⁿa_n converges. But, like I said before, do I know that a1 >= a2 >= ... >= an because the series converges at R?

Dick

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