Convergent non-monotone sequences

  • Thread starter Easty
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  • #1
Easty
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Homework Statement



Let C=[tex]\bigcup[/tex]n=1[tex]\infty[/tex]Cn where Cn=[1/n,3-(1/n)]
a) Find C in its simplest form.
b)Give a non-monotone sequence in C converging to 0.

Homework Equations




The Attempt at a Solution


For part a) i get C=[0,3]. Is this correct? I am not sure as to wether 0 and 3 are contained in the set though. Should it be C=(0, 3)?

As for part b) I am not really sure here. I thought one such sequence might be
(1, 2, 1/2, 1/3, 1/4, 1/5,...). So the tail converges to zero and the first 2 terms mean it is non-monotone.
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Answers and Replies

  • #2
tiny-tim
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Hi Easty! :smile:
For part a) i get C=[0,3]. Is this correct? I am not sure as to wether 0 and 3 are contained in the set though.

Well, by definition of union, 0 is only in C if it's in one of the Cns … is it? :wink:
As for part b) I am not really sure here. I thought one such sequence might be
(1, 2, 1/2, 1/3, 1/4, 1/5,...). So the tail converges to zero and the first 2 terms mean it is non-monotone.

hmm … seems a daft question :rolleyes:

but I suspect they want it to be non-monotone wherever you start.
 
  • #3
Easty
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Ok then so it must be C=(0, 3).

As for part b would this sequence work:
(1/n*sin(n)).
It should converge by the absolute convergence theorem i think.
I'd appreciate any comments or criticisms.
thanks
 
  • #4
tiny-tim
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Ok then so it must be C=(0, 3).

Yup! :biggrin:
As for part b would this sequence work:
(1/n*sin(n)).
It should converge by the absolute convergence theorem i think.

Yes, almost any daft sequence works! :smile:
 

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