Proof on Sequences: Sum of a convergent and divergent diverges


by Heute
Tags: convergent, divergent, diverges, proof, sequences
Heute
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#1
Oct10-11, 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.
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SammyS
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#2
Oct10-11, 06:35 PM
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That's the general idea !

Rewrite [itex]\left|a_n+b_n-L\right|[/itex] as [itex]\left|a_n-M+b_n-(L-M)\right|\,,[/itex] then use the triangle inequality.
Heute
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#3
Oct10-11, 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] - (L-M)| < 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] - (L-M)| < ε
which implies [itex]b_{n}[/itex] converges, which is a contradiction
Therefore, [itex]a_{n}[/itex]+[itex]b_{n}[/itex] diverges

SammyS
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#4
Oct10-11, 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]\left|a_n+b_n-L-a_n+M\right|\le\left|a_n+b_n-L\right|+\left|-a_n+M\right|<\varepsilon[/itex]


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