sequence proof

[happyg1]

Hi,
Here is the question:
Prove that if the sequence {s} has no convergent subsequence then {|s|} diverges to infinity.

To me, this seems so easy, but I'm having a really hard time putting it down in a rigorous manner.
My thoughts are:
every convergent sequence has a convergent subsequence (theorem in the book), so if there is no convergent subsequence, then the sequence itsself cannot converge either.
So the absolute value won't converge.

I tried to do it with contradiction as follows:
Suppose that |s| converges. Then it has a convergent subsequence. Since |s| is also a subsequence of {s}, the convergent subsequence of |s| lies inside {s}. So {s} has a convergent subsequence. Contradiction.

Any clarification will be greatly appreciated.
CC

[CarlB]

I tried to do it with contradiction as follows:

Suppose that |s| converges. Then it has a convergent subsequence. Since |s| is also a subsequence of {s}, the convergent subsequence of |s| lies inside {s}. So {s} has a convergent subsequence. Contradiction.

There is a problem with your logic in that {|s|} is not necessarily a subsequence of {s}. This is true even for convergent series like:

s_n = \frac{(-1)^n}{n}

Carl

[happyg1]

Ok,
I see that. Can I say that the elements a convergent subsequence of |s| also are members of {s}? The convergent subsequence of |s| will be positive numbers converging to L, so {s} will either have the same exact subsequence of positive terms, or it'll be negative terms, but then will converge to -L

[CarlB]

happyg1,

Was the original problem "Prove that if the sequence {s} has no convergent subsequence then {|s|} diverges to infinity.", or was it instead "Prove that if the sequence {s} has no convergent subsequence then \sum |s| diverges to infinity."

The reason I ask this is because I'm not really sure what "diverges to infinity" means.

If the meaning is to say that given a number L, no matter how far down the series you look, you can find an element larger in absolute value than L, then I can see your terminology working.

The whole thing smells of recourse to an argument based on the fact that any sequence that stays inside a closed subset of the real line has a convergent subsequence.

Carl

[happyg1]

Prove that if the sequence {s} has no convergent subsequence then {|s|} diverges to infinity.

that is the question as printed in my book.

[CarlB]

Then I think you will need to make an argument based on a fact that I vaguely recall to the effect that any sequence to a compact space has a subsequence that converges to an element of the compact space.

Hey, it's been about 30 years since I studied this, and this is the limit of what I can help.

Carl