Dean Hanafy	Posted on Monday, 16 February, 2004 - 05:50 pm: Can someone give me a hint on how to do the following part question: Prove that an increasing unbounded sequence tends to infinity. Thanks Deano
Francis Woodhouse	Posted on Monday, 16 February, 2004 - 07:33 pm: I'm guessing here, but you could try something with an arbitrary bound and the nth value of the sequence that increases it above that bound. That is, if we have some A, then we can find an n(A) such that an(A) > A, since (an ) is unbounded and increasing.
William Hall	Posted on Monday, 16 February, 2004 - 09:07 pm: What Francis says is pretty much the crux of the proof. What you need to do now is to use the idea to prove the theorem. Hint: Try proof by contradiction. Bill
Dean Hanafy	Posted on Monday, 16 February, 2004 - 10:37 pm: Can you give me a little further hint if possible? I mean does the contradiction lie in assuming that an unbounded sequence tends to a value, say S, and then arriving at a value greater than S? I never no where to start with analysis proofs! I had an alternative but i'm not sure whether i have prove what needs to be proved! Could someone please tell me if the following suffices: Assume that the unbounded increasing sequence tends to a value a. But by the monotone convergence theorem, only if a sequence is bounded from above does the sequence converge to the supremum. But since the sequence is unbounded, it doesn't have a supremum. Since it doesn't have a supremum, it doesn't converge to some value a. And since it is increasing, each subsequent term is greater that the previous. Therefore it must tend to infinity. Is that ok? Please tell me how i can refine my argument, if it is valid(!) or give me any hints on how to arrive at a contradiction Thanks Deano
Michael McLoughlin	Posted on Monday, 16 February, 2004 - 10:40 pm: If what I am about to say is stupid then please tell me to shut up, but here goes... Surely an 'unbounded sequence' which does not tend to infinity is not an unbounded seqence.
Michael Doré	Posted on Monday, 16 February, 2004 - 10:45 pm: I think that looks OK. (There are one or two bits I don't quite follow, but I think your idea can be made to work.) But it's worth noting that you don't really need proof by contradiction. Nor do you need to use the completeness axiom for the reals. Indeed the result is true if we work in the rationals instead of the reals. All you need is the definition of 'tends to infinity'. By definition a sequence an tends to infinity iff for every K there exists natural N such that for all n > N we have an > K. So let K be arbitrary. Since the sequence is unbounded there exists N such that aN > K. What can you say about an for all n > N? Michael PS. Michael - there are unbounded sequences which do not tend to infinity. For example set an = n for n odd and an = 0 for n even.
Julian Pulman	Posted on Monday, 16 February, 2004 - 10:45 pm: Michael, consider 1,0,2,0,3,0,...,n,0,... This is unbounded, but does not tend to infinity.
Francis Woodhouse	Posted on Monday, 16 February, 2004 - 10:59 pm: Michael: An unbounded MONOTONIC sequence which does not tend to infinity (negative or positive) is not an unbounded sequence. It's important to include the monotonic part there (monotonic means either non-decreasing or non-increasing).
Michael McLoughlin	Posted on Tuesday, 17 February, 2004 - 12:02 am: Thank you, I think I understand now. However, considering what Francis said in the last post about monotonic sequences and then refering to the wording in the question which said 'increasing unbounded sequence' (a sequence is monotonic if it is increasing), is it not clearly the case that this would tend to infinity. All I am trying to get my head round is why the statement needs proof; to me, it reads like a tautology. Anyway, am I still missing somthing? It will not be the first time I have been shouted down on this forum for lacking rigour.
Francis Woodhouse	Posted on Tuesday, 17 February, 2004 - 12:22 am: It is simply due to the rigour of pure mathematics. Assume nothing; prove everything. You have to show through logical reasoning exactly why the statement is true. If you don't like this level of proof, then I'm sure you'll be a perfect applied mathematician, but pure mathematicians go into excruciating detail over things. Try getting your head around some of the $\epsilon -\delta$ arguments for continuity and suchlike for the first time - it hurts your brain, to begin with!
Michael Doré	Posted on Tuesday, 17 February, 2004 - 12:25 am: Well it is clearly true - but I don't think it is a tautology (in the sense you're using the word). The point is you need to use the definitions of all the terms. We're asked: Prove that an increasing unbounded sequence tends to infinity. An increasing sequence an is one such that an < = an+1 for all n. An unbounded sequence is one such that for every K there exists n such that an > K. And a sequence tends to infinity iff for every K' there exists N such that an > K' for all n > N. So putting all that in, what we're asked is: If an is a sequence, an < = an+1 for all n, and for every K there exists n such that an > K, prove that for every K' there exists N such that an > K' for all n > N. Is that a tautology? I agree it is easy to prove... but surely there is actually something to prove here? Michael
Julian Pulman	Posted on Tuesday, 17 February, 2004 - 12:32 am: The reason why you cannot simply 'claim' it's obvious is because of the need for rigour. In real analysis we tend to use an 'epsilon' approach to things, so if a sequence an tends to a limit a we write down Why do we need to get so pedantic and make sure it's all translated in mathematical notation? It's because there are problems which aren't so intuitive (like this problem) and which a common sensical approach will not suffice. A great example of this is the Intermediate Value Theorem, which can be proved entirely from the axiom of analysis (that an increasing, bounded sequence tends to a limit) - I remember Professor Korner in my 1A Analysis lectures stressing again and again that this is NOT an obvious theorem. I think that the reason you don't think the statement needs a proof IS actually the proof itself, if you order your thoughts into a string of arguments - it should probably mirror the analysis proof. (e.g. compare with a sequence such as 3,3.1,3.14,3.141,3.1415,3.14159,..., which is increasing, but bounded.)
Julian Pulman	Posted on Tuesday, 17 February, 2004 - 12:33 am: ...I think I should probably start checking for double postings in future.