Finding Partial Limits of a Sequence: A Homework Challenge

estro
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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


a_1=0

a_{2n}=\frac {a_{2n-1}} {3}

a_{2n+1} = 1/3 + a_{2n}

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.
 
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estro said:

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


a_1=0

a_{2n}=\frac {a_{2n-1}} {3}

a_{2n+1} = 1/3 + 2_{2n}
In the equation above do you mean a2n+1 = 1/3 + 22n?
estro said:

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.
 
Sorry, have fixed it in my first post.
 
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?
 
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
 
I have figured it out, thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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