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Proof on Sequences: Sum of a convergent and divergent diverges 
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#1
Oct1011, 06:13 PM

P: 25

1. The problem statement, all variables and given/known data
Prove if sequence [itex]a_{n}[/itex] converges and sequence [itex]b_{n}[/itex] diverges, then the sequence [itex]a_{n}[/itex]+[itex]b_{n}[/itex] also diverges. 2. Relevant equations 3. The attempt at a solution My professor recommended a proof by contradiction. That is, suppose [itex]a_{n}[/itex]+[itex]b_{n}[/itex] does converge. Then, for every ε > 0, there exists a natural number [itex]N_{1}[/itex] so that n > [itex]N_{1}[/itex] implies [itex]a_{n}[/itex]+[itex]b_{n}[/itex]  L< ε We already know there exists [itex]N_{2}[/itex] so that n > [itex]N_{2}[/itex] implies [itex]a_{n}[/itex]  M < ε. So let N = max{[itex]N_{1}[/itex], [itex]N_{2}[/itex]}. Then n > N means we know [itex]a_{n}[/itex] is "very close" to M. My purpose in this is to try and show that this implies [itex]b_{n}[/itex] has a limit (that is, it converges) providing a contradiction. However, I'm not sure how to go about this. 


#2
Oct1011, 06:35 PM

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That's the general idea !
Rewrite [itex]\lefta_n+b_nL\right[/itex] as [itex]\lefta_nM+b_n(LM)\right\,,[/itex] then use the triangle inequality. 


#3
Oct1011, 11:01 PM

P: 25

Ah! I think I've got it.
Proof: Assume [itex]a_{n}[/itex] is convergent and [itex]b_{n}[/itex] is divergent. Now suppose that [itex]a_{n}[/itex]+[itex]b_{n}[/itex] is convergent. Then [for every ε > 0 there exists a natural number [itex]N_{1}[/itex] so that n > [itex]N_{1}[/itex] implies [itex]a_{n}[/itex]+[itex]b_{n}[/itex]  L< ε/2 We know by our assumption that there also exists natural number [itex]N_{2}[/itex] so that n > [itex]N_{2}[/itex] implies [itex]a_{n}[/itex]M < ε/2 Now let N = max{[itex]N_{1}[/itex],[itex]N_{2}[/itex]} Then n > N implies [itex]a_{n}[/itex]+[itex]b_{n}[/itex]  L< ε/2 and [itex]a_{n}[/itex]M < ε/2 So [itex]a_{n}[/itex]+[itex]b_{n}[/itex]  L+ ([itex]a_{n}[/itex]M) < ε and [itex]a_{n}[/itex]+[itex]b_{n}[/itex]  [itex]a_{n}[/itex]  (LM) < ε (by the triangle inequality) But [itex]a_{n}[/itex]+[itex]b_{n}[/itex]  [itex]a_{n}[/itex] = [itex]b_{n}[/itex] so this last statement implies [itex]b_{n}[/itex]  (LM) < ε which implies [itex]b_{n}[/itex] converges, which is a contradiction Therefore, [itex]a_{n}[/itex]+[itex]b_{n}[/itex] diverges 


#4
Oct1011, 11:32 PM

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Proof on Sequences: Sum of a convergent and divergent diverges
Make it clear that by the triangle ineq. [itex]\lefta_n+b_nLa_n+M\right\le\lefta_n+b_nL\right+\lefta_n+M\right<\varepsilon[/itex]



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