ANALYSIS Q i need a hand with please!!

Nappy

Given a convergent sequence (a)n how can I prove that the sequence has an infinite number of subsequences??

Ok my attempt so far, dont really know where to start with it, but

We know a convergent sequence converges to a limit, so for instance
for the series
(a)n = 1/n
(a)n = 1, .5, .3333, .25, .2,.........., 1/n
this sequence is monotone decreasing, any sub sequence would also have the same property, for example
sub1(a)= .5, .25,.2
this ssubsequence is still converging to the limit as aall the values still hold their positions in the sequence. As this sequence can go on infinitely as n tends to infinity, it can therefore have infinite subsequences.

This makes sense to me but I dont think its a concrete proof?? Its an assignment and im ttrying to spruce it up, there seems to be far too much writing for a maths assignment.

Any help would be much appreciated.

Cheers

Nappy

MathsManiac

Are you sure you have the question right? Every infinite sequence (convergent or otherwise) has an infinite number of subsequences. After all, the sequence that starts at any particular term and simply continues along the original sequence is a subsequence. And there are clearly infinitely many of those. (It's also fairly trivial to prove that if the original sequence converges, then so too do all of these.)

Nappy

Q: Let (a)n be a convergent sequence. Prove that (a)n has an infinite number of different subsequences.

How would you go about proving it?? It obvious but I just cant put it into mathematical notation..

Can you use proof by contradiction here?

Fremen

An infinite sequence can be indexed by the natural numbers {1,2,3,4,...}. Suppose S = a1, a2, a3, a4,....
The set of natural numbers is infinite (countable). We just need to show that the set of natural numbers has an infinite number of subsets which are also infinite.

We do this by explicit construction: Let
N = {1,2,3,4,...}
2N = {2,4,6,8,...}
3N= {3,6,9,12,...}

kN is infinite for each k, and there are an infinite number of distinct k's. This is all we need.

Seems like a more sensible question would be to show that if S converges to a limit, S has an infinite number of infinite subsequences, each of which converge to the same limit.

Edit: Actually, you can just define a subsequence S_n as S with the n-th element removed. Clearly you can generate an infinite number of different subsequences this way.

rjt

What about s_n = 0? :P

MathsManiac

rjt wrote: »

What about s_n = 0? :P

What about it?

(It would obviously not, in general, be a subsequence of the original sequence, if that's what you meant.)

rjt

MathsManiac wrote: »

What about it?

(It would obviously not, in general, be a subsequence of the original sequence, if that's what you meant.)

Well, my interpretation of the question is: show that any convergent sequence s_n has infinitely many different subsequences. If we remove the word "different" then, as you pointed out earlier, the question doesn't make much sense. But s_n = 0 is convergent, and there aren't infinitely many different subsequences.

MathsManiac

Now I see what you mean. The OP used a(n) for the original sequence and Fremen used s(n) for the subsequences, so I hope you can see why I thought you were referring to the subsequences and not the original sequence when you said s_n=0.

Anyway, back to the point: my first post had nothing to do with providing a whole pile of identical sequences. (Clearly the case where all terms are equal is a special case in which all subsequences are identical.)

In general, however, the subsequences I pointed out in my first post are all distinct. That is, defining b_r(n) := a(n+r) gives infinitely many subsequences. (That is, the first term of sequence b_r is the (r+1)th term of sequence a). Each different value of r gives a distinct subsequence, (provided the sequence doesn't eventually settle to a constant).

You have of course pointed out a flaw in the original question. If the sequences have to be distinct in order to say that there are infinitely many of them (as seems eminently reaonable), then the statement is false in any case where the sequence is finitely non-zero.

rjt

MathsManiac wrote: »

Now I see what you mean. The OP used a(n) for the original sequence and Fremen used s(n) for the subsequences, so I hope you can see why I thought you were referring to the subsequences and not the original sequence when you said s_n=0.

Anyway, back to the point: my first post had nothing to do with providing a whole pile of identical sequences. (Clearly the case where all terms are equal is a special case in which all subsequences are identical.)

In general, however, the subsequences I pointed out in my first post are all distinct. That is, defining b_r(n) := a(n+r) gives infinitely many subsequences. (That is, the first term of sequence b_r is the (r+1)th term of sequence a). Each different value of r gives a distinct subsequence, (provided the sequence doesn't eventually settle to a constant).

You have of course pointed out a flaw in the original question. If the sequences have to be distinct in order to say that there are infinitely many of them (as seems eminently reaonable), then the statement is false in any case where the sequence is finitely non-zero.

Yeah, sorry. In hindsight I wasn't very clear!