Finding Partial Limits of a Sequence: A Homework Challenge

[estro]

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.

 

[Mark44]

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 a_2n+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.

 

[estro]

Sorry, have fixed it in my first post.

 

[Mark44]

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?

 

[estro]

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

 

[estro]

I have figured it out, thanks.