1. The problem statement, all variables and given/known data Suppose that [tex]\sum[/tex]a_{n}x^{n} 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_{n}R^{n} = M. At -R, the series is the following: [tex]\sum[/tex]a_{n}(-R)^{n} = [tex]\sum[/tex](-1)^{n}a_{n}R^{n}. 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.
Thanks! So since [tex]\sum[/tex]a_{n}R^{n} converges, and a_{n}(-R)^{n} <= a_{n}R^{n} for all n, then [tex]\sum[/tex]a_{n}(-R)^{n} converges.
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]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_{n}R^{n}, 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?
Nothing wrong with that. Unfortunately, I wasn't thinking of that comparison test. Hence the panic attack. Carry on.
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?
If a1 >= a2 >= ... >= an >= ... >= 0 and lim(an) = 0, then the alternating series [tex]\sum[/tex](-1)^{n}a_{n} converges. But, like I said before, do I know that a1 >= a2 >= ... >= an because the series converges at R?
All of terms a_n*R^n are positive and it's convergent. The series is absolutely convergent. Nothing can go wrong here.