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Partial limit

  • Thread starter estro
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  • #1
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Homework Statement


I have to find all the partial limits {I hope this is how this term named in English} of a sequences


Homework Equations


[tex]a_1=0[/tex]

[tex]a_{2n}=\frac {a_{2n-1}} {3}[/tex]

[tex] a_{2n+1} = 1/3 + a_{2n}[/tex]



The Attempt at a Solution


I have tried to prove first that sequences of all the even terms converges due to fact that sequence is monotonic and have a suprimum, but have failed to prove it.
Another problem is that subsequences of odd term is non monotonic, but I also can't use the Cantor's Lemma.

Could you please suggest how to approach this problem?
Thanks.
 
Last edited:

Answers and Replies

  • #2
33,496
5,188

Homework Statement


I have to find all the partial limits {I hope this is how this term named in English} of a sequences


Homework Equations


[tex]a_1=0[/tex]

[tex]a_{2n}=\frac {a_{2n-1}} {3}[/tex]

[tex] a_{2n+1} = 1/3 + 2_{2n}[/tex]
In the equation above do you mean a2n+1 = 1/3 + 22n?

The Attempt at a Solution


I have tried to prove first that sequences of all the even terms converges due to fact that sequence is monotonic and have a suprimum, but have failed to prove it.
Another problem is that subsequences of odd term is non monotonic, but I also can't use the Cantor's Lemma.

Could you please suggest how to approach this problem?
Thanks.
 
  • #3
241
0
Sorry, have fixed it in my first post.
 
  • #4
33,496
5,188
You're looking at the two subsequences: one with the odd-index terms and the other with the even-index terms. Have you calculated the first dozen or so terms of your sequence?
 
  • #5
241
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I've calculated again some terms of the sequence and found out that I did a mistake in my previous calculation as both subsequences seem to be monotonic, but I can't find a way to prove that the sequences have suprimums.

0, 0, 81/243, 27/243, 108/243, 36/243, 36/243, 117/243, 39/243, 120/243, 40/243, 121/243
 
  • #6
241
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I have figured it out, thanks.
 

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