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Prove series converges

  • Thread starter JG89
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  • #1
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Homework Statement



If a_1 + a_2 + ... is an infinite series converging to A, and b1, b2, ... is an infinite sequence that is bounded and monotonic, prove that (a_1)(b_1) + (a_2)(b_2) + ... converges


Homework Equations





The Attempt at a Solution



I don't really know where to start...all I can say is that if a_1 + a_2 + ... converges, then a_n approaches 0 as n goes to infinity, and so (a_n)(b_n) also has a limit of 0, since b_n converges to some finite value.
 

Answers and Replies

  • #2
Office_Shredder
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You know that bn is bounded and monotonic. Have you tried using that? Given a general number b, you know
[tex] \sum a_n b [/tex]
converges right? Now try to compare that to the sequence

[tex] \sum a_n b_n[/tex] choosing b such that |bn| < b for all n
 
  • #3
tiny-tim
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Hi JG89! :smile:
… since b_n converges to some finite value.
Yes :approve: … concentrate on that value (call it b) …

then use deltas and epsilons. :wink:
 
  • #4
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I have an intuitive idea of what's going on, but I'm having a hard time fleshing it out into an epsilon argument.

First off, since a_1 + a_2 + ... converges then b(a_1 + a_2 + ...) converges. I know that for large enough n, |(a_n)(b_n)| gets 'really' close to |(a_n)b|, since b is the limit of b_n. Since the b_n are monotonic, then as n increases the |(a_n)(b_n)| gets even closer to |(a_n)b|. I can in fact make this difference as small as I please, provided n is taken large enough and so it seems that after a certain n, the difference between (a_n)(b_n) and (a_n)b will become "negligible" and since (a_n)b is Cauchy, then (a_n)(b_n) is also Cauchy.
 
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