How can I prove or disprove that if a and a + b are divergent sequences then b is a divergent sequence?

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Prove that if an and an + bn are divergent sequences then bn is a divergent sequence.




real-analysis sequences-and-series




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    It doesn't have to be: For $a_n=n$, and $b_n=0$. You have that both $a_n=n$ and $a_n+b_n=a_n=n$ are divergent, while $b_n=0$ is convergent.
    – user574889
    Jul 20 at 1:48











  • Thanks cactus. Got it!
    – Joshua
    Jul 20 at 1:52














up vote
-5
down vote

favorite












Prove that if an and an + bn are divergent sequences then bn is a divergent sequence.




real-analysis sequences-and-series




share|cite|improve this question

















  • 2




    It doesn't have to be: For $a_n=n$, and $b_n=0$. You have that both $a_n=n$ and $a_n+b_n=a_n=n$ are divergent, while $b_n=0$ is convergent.
    – user574889
    Jul 20 at 1:48











  • Thanks cactus. Got it!
    – Joshua
    Jul 20 at 1:52












up vote
-5
down vote

favorite









up vote
-5
down vote

favorite











Prove that if an and an + bn are divergent sequences then bn is a divergent sequence.




real-analysis sequences-and-series




share|cite|improve this question













Prove that if an and an + bn are divergent sequences then bn is a divergent sequence.





real-analysis sequences-and-series





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 20 at 2:02
























asked Jul 20 at 1:47









Joshua

12




12







  • 2




    It doesn't have to be: For $a_n=n$, and $b_n=0$. You have that both $a_n=n$ and $a_n+b_n=a_n=n$ are divergent, while $b_n=0$ is convergent.
    – user574889
    Jul 20 at 1:48











  • Thanks cactus. Got it!
    – Joshua
    Jul 20 at 1:52












  • 2




    It doesn't have to be: For $a_n=n$, and $b_n=0$. You have that both $a_n=n$ and $a_n+b_n=a_n=n$ are divergent, while $b_n=0$ is convergent.
    – user574889
    Jul 20 at 1:48











  • Thanks cactus. Got it!
    – Joshua
    Jul 20 at 1:52







2




2




It doesn't have to be: For $a_n=n$, and $b_n=0$. You have that both $a_n=n$ and $a_n+b_n=a_n=n$ are divergent, while $b_n=0$ is convergent.
– user574889
Jul 20 at 1:48





It doesn't have to be: For $a_n=n$, and $b_n=0$. You have that both $a_n=n$ and $a_n+b_n=a_n=n$ are divergent, while $b_n=0$ is convergent.
– user574889
Jul 20 at 1:48













Thanks cactus. Got it!
– Joshua
Jul 20 at 1:52




Thanks cactus. Got it!
– Joshua
Jul 20 at 1:52










1 Answer
1






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oldest

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up vote
3
down vote













Let $A_n=n $ and $B_n = 1/n$. Then $A_n$ diverges and $A_n+B_n$ diverges, yet $B_n$ converges - a counterexample to your proposal.






share|cite|improve this answer























  • Fixed it. Thanks.
    – DavidS
    Jul 20 at 1:56






  • 2




    $A_n = 1$ isn't divergent.
    – Brian Yao
    Jul 20 at 2:39






  • 2




    @Squirtle You’re going to skip upvoting a quality answer from a new member because . . . it’s “too” good?
    – Chase Ryan Taylor
    Jul 20 at 3:11










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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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up vote
3
down vote













Let $A_n=n $ and $B_n = 1/n$. Then $A_n$ diverges and $A_n+B_n$ diverges, yet $B_n$ converges - a counterexample to your proposal.






share|cite|improve this answer























  • Fixed it. Thanks.
    – DavidS
    Jul 20 at 1:56






  • 2




    $A_n = 1$ isn't divergent.
    – Brian Yao
    Jul 20 at 2:39






  • 2




    @Squirtle You’re going to skip upvoting a quality answer from a new member because . . . it’s “too” good?
    – Chase Ryan Taylor
    Jul 20 at 3:11














up vote
3
down vote













Let $A_n=n $ and $B_n = 1/n$. Then $A_n$ diverges and $A_n+B_n$ diverges, yet $B_n$ converges - a counterexample to your proposal.






share|cite|improve this answer























  • Fixed it. Thanks.
    – DavidS
    Jul 20 at 1:56






  • 2




    $A_n = 1$ isn't divergent.
    – Brian Yao
    Jul 20 at 2:39






  • 2




    @Squirtle You’re going to skip upvoting a quality answer from a new member because . . . it’s “too” good?
    – Chase Ryan Taylor
    Jul 20 at 3:11












up vote
3
down vote










up vote
3
down vote









Let $A_n=n $ and $B_n = 1/n$. Then $A_n$ diverges and $A_n+B_n$ diverges, yet $B_n$ converges - a counterexample to your proposal.






share|cite|improve this answer















Let $A_n=n $ and $B_n = 1/n$. Then $A_n$ diverges and $A_n+B_n$ diverges, yet $B_n$ converges - a counterexample to your proposal.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 20 at 1:56


























answered Jul 20 at 1:53









DavidS

858




858











  • Fixed it. Thanks.
    – DavidS
    Jul 20 at 1:56






  • 2




    $A_n = 1$ isn't divergent.
    – Brian Yao
    Jul 20 at 2:39






  • 2




    @Squirtle You’re going to skip upvoting a quality answer from a new member because . . . it’s “too” good?
    – Chase Ryan Taylor
    Jul 20 at 3:11
















  • Fixed it. Thanks.
    – DavidS
    Jul 20 at 1:56






  • 2




    $A_n = 1$ isn't divergent.
    – Brian Yao
    Jul 20 at 2:39






  • 2




    @Squirtle You’re going to skip upvoting a quality answer from a new member because . . . it’s “too” good?
    – Chase Ryan Taylor
    Jul 20 at 3:11















Fixed it. Thanks.
– DavidS
Jul 20 at 1:56




Fixed it. Thanks.
– DavidS
Jul 20 at 1:56




2




2




$A_n = 1$ isn't divergent.
– Brian Yao
Jul 20 at 2:39




$A_n = 1$ isn't divergent.
– Brian Yao
Jul 20 at 2:39




2




2




@Squirtle You’re going to skip upvoting a quality answer from a new member because . . . it’s “too” good?
– Chase Ryan Taylor
Jul 20 at 3:11




@Squirtle You’re going to skip upvoting a quality answer from a new member because . . . it’s “too” good?
– Chase Ryan Taylor
Jul 20 at 3:11












 

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