# Sequence: nondecreasing, bounded above, prove s_n < L

1. Oct 6, 2009

### tarheelborn

1. The problem statement, all variables and given/known data
If {s_n} is nondecreasing and bounded above, and L = lim s_n, prove that s_n <= L.

2. Relevant equations

3. The attempt at a solution
This is one of those proofs that seems, to me, to be obvious from the proven theorem that states that the limit of a sequence is equal to the least upper bound. But I doubt my professor would both with it if it were obvious. Please help!

2. Oct 6, 2009

### Dick

Mmm. Proof by contradiction? Assume there is an s_k such that s_k>L. Now what?

3. Oct 6, 2009

### tarheelborn

If s_k > L, then L is not a least upper bound for s_k and, hence, L is not the limit for s_k. This is a contradiction, so L < s_k? How about that?

4. Oct 6, 2009

### Dick

I guess that's ok, if you know that a bounded nondecreasing sequence has a limit that is the LUB of the sequence. You could maybe be a little more direct and define s_k-L=e>0. Then since the sequence is nondecreasing, s_i-L>e for all i>k. L is not the limit.

Last edited: Oct 6, 2009
5. Oct 6, 2009

### tarheelborn

We did just prove a theorem that the least upper bound = the limit of a nondecreasing, convergent sequence, but... it does seem like we always come back to the epsilon proof. Along that line:

If e > 0, we need to find N such that s_k - L < e. But since s_k is nondecreasing, there is i > 0 such that s_i - L >= e, so L is not the limit. But this contradicts our hypothesis, so in fact, s_n < L. End.

This does make sense to me, although I wouldn't have come up with a different index, i. e., "i", unless you had shown me that.

6. Oct 6, 2009

### Dick

You've got the idea, but the use of the indices is still sloppy. If there were AN s_k>L and e=s_k-L then s_i-L>=s_k-L=e for ALL i>k. If the lim(s)=L then there is AN N such that |s_i-L|<e for ALL i>N. Clearly the two conflict for i>max(N,k). Be sure and distinguish between indices like 'k' and 'N' which identify a single element of the sequence and indices like 'i' which hold for a whole range.

7. Oct 6, 2009

### tarheelborn

I am very confused by all the different indices, but I have gotten this far: Assume there is an s_k such that s_k > L. Then for e > 0, s_k - L = e. But since the sequence is nondecreasing, there is also an s_i - L > s_k - L = e. If L is the limit, then there exists an N such that |s_i - L| < e for all i >=N. But

I can't quite make the conclusion. It seems to me as if finding s_i - L > s_k - L should be a conclusion, but that doesn't seem quite complete. I really appreciate your help.

8. Oct 6, 2009

### tarheelborn

And, hitting the middle between my original thought and your suggestions, please look at this version:

Suppose that there is an s_k such that s_k > L. Then for e e> 0, s_k - L >= e. But {s_n} is nondecreasing and bounded above, so s_n has a least upper bound, say M. By theorem ( a nondecreasing sequence which is bounded above is convergent, that is, L = M), lim s_n = M, so we have a contradiction. That is, if M is the least upper bound for s_n, then M = L, so that s_k cannot be greater than L. Therefore, s_k <= L implies that s_n <= L. End of proof.

9. Oct 6, 2009

### Dick

i) Don't just pick any epsilon. Pick e=s_k-L. ii) Since the sequence is nondecreasing s_i-L>=s_k-L=e FOR ALL i>k. A single value of i isn't enough. If you just had a single value of i such that s_i-L>=e, you might be able pick N bigger that that value of i. But if it's true for ALL i>k then there clearly is no such N.

10. Oct 6, 2009

### Dick

Actually, your original proof was fine. You are just making it more complicated without making it better. By your theorem you know the limit, L, is the LUB of the s_k. If s_k>L then L can't be the LUB. That's a contradiction, so s_k is not greater than L.

11. Oct 6, 2009

### tarheelborn

Thanks so much! You cannot imagine how that helps my confidence. Have a great evening.

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