
#1
Apr2908, 10:08 AM

P: 159

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. 



#2
Apr2908, 10:14 AM

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




#3
Apr2908, 10:30 AM

P: 159

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.




#4
Apr2908, 10:32 AM

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Real Analysis  Radius of Convergence
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?




#5
Apr2908, 11:19 AM

P: 159

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? 



#6
Apr2908, 11:27 AM

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




#7
Apr2908, 11:28 AM

P: 159

Ok! Thanks once again for your help.




#8
Apr2908, 12:15 PM

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P: 38,881

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? 



#9
Apr2908, 12:37 PM

P: 159





#10
Apr2908, 01:30 PM

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