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Math Help - Definition of sequence to infinity

  1. #1
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    Definition of sequence to infinity

    Hey

    Can anyone explain what A and N are in the definition for a limit of infinity in a sequence:

    a_{n} > A ~~for~all~~ n > N

    I've reread he page over and over again but it's not sinking in. I probably just need to sleep on the idea!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by aceband View Post
    Hey

    Can anyone explain what A and N are in the definition for a limit of infinity in a sequence:

    a_{n} > A ~~for~all~~ n > N

    I've reread he page over and over again but it's not sinking in. I probably just need to sleep on the idea!
    So, what this definition is saying is that, if a_n repesents a sequence of numbers, then this sequence grows without bound if each member of the sequence is greater than some arbitrarily Large number A whenever n>N.
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  3. #3
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    Thanks for the quick reply!

    I think i struggle to understand because i don't see how it would be used to show an infiinte upper bound (is that correct term).

    Where would A and N come from?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by aceband View Post
    Thanks for the quick reply!

    I think i struggle to understand because i don't see how it would be used to show an infiinte upper bound (is that correct term).

    Where would A and N come from?
    The thing here to realize is that A is chosen to be as large as you wish. So, the proof of an unbounded sequence lies in showing that FOR ANY arbitrarily large numbers A and N (assuming a_N\geq{A}), there exists an n such that a_n>A whenever n>N
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aceband View Post
    Hey

    Can anyone explain what A and N are in the definition for a limit of infinity in a sequence:

    a_{n} > A ~~for~all~~ n > N

    I've reread he page over and over again but it's not sinking in. I probably just need to sleep on the idea!
    A is any finite number. Any number. as big as you want. It is saying that no matter what finite number you pick, all the terms of the sequence will exceed it eventually, after you pass some term a_n.

    An example, the sequence \{ a_n \} = n goes to infinity as n goes to infinity.

    How do we know? well, if some arbitrary number A is given, all we have to do is choose any N > \lceil A \rceil (the smallest integer greater than A) and then if n > N, a_n > A (that is, all terms after the Nth term will be bigger than the number A.

    Example, say someone choose A = 14.7, we can pick, N = 16, and so a_{16},~a_{17},~a_{18}, \dots will all be bigger than A, these will be the integers 16, 17, 18, ...

    got it? No matter what number you pick, our sequence will surpass it eventually, in fact, every term of our sequence from some point on will pass it. the fact that this will happen as you keep picking bigger and bigger numbers, means the terms of our sequence can get as big as you want. in technical terms, the sequence tends to infinity
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  6. #6
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    Okey doke,

    See if i can get this:
    Let a_{n} = n^{2}

    Which should tend to infinity.
    To show it i need to show that there exists an n where ' a_{n} > A' and 'n > N'

    Is the first step to put n = A?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aceband View Post
    Okey doke,

    See if i can get this:
    Let a_{n} = n^{2}

    Which should tend to infinity.
    To show it i need to show that there exists an n where ' a_{n} > A' and 'n > N'

    Is the first step to put n = A?
    well, that would work actually--except if 0 < A < 1.

    i thought getting the idea was the important thing here. do you get the idea?

    to apply it, we have to be a little more technical and stick to the definition closely. you would prove this as follows:




    Let A > 0 be given. Choose N = \text{max} \left \{ 1, \sqrt A \right \} (this is to deal with the case where 0 < A < 1, if it is in that range, we simply choose N = 1)

    Then, n > N implies a_n = n^2 > N^2 = \sqrt A ^2 = |A| \ge A.

    So \lim_{n \to \infty} a_n = \infty by definition.
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  8. #8
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    Hmm i think this is just too much for me to grasp. I can see it hsouldn't be so hard to understand but it just wont click!
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aceband View Post
    Hmm i think this is just too much for me to grasp. I can see it hsouldn't be so hard to understand but it just wont click!
    it's hard for everyone when just learning this. don't be discouraged. how about having a crack at this one...

    Show that \lim_{n \to \infty}(2n + 1) = \infty
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  10. #10
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    Thanks for the words,

    Ok lets see then.

    Is it along these lines:
    <br /> <br />
a_{n} = 2n + 1 \\
    <br />
a_{n} > A \\
    <br />
So~~2n+1 > A ~~for~all~~ n > N<br /> <br />


    Before i hash it up any more do i need ot rearrangem get n on it's own at all?
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aceband View Post
    Thanks for the words,

    Ok lets see then.

    Is it along these lines:
    <br /> <br />
a_{n} = 2n + 1 \\
    <br />
a_{n} > A \\
    <br />
So~~2n+1 > A ~~for~all~~ n > N<br /> <br />

    Before i hash it up any more do i need ot rearrangem get n on it's own at all?
    you can just solve for n as you would in a regular equation. then you will know what to pick N to be. (By the way, nothing you do so far would show up in your proof. this is just the rough work to know what you'll put into it.
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  12. #12
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    So n = 1/2(A - 1) which in turn is greater than N?
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  13. #13
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aceband View Post
    So n = 1/2(A - 1) which in turn is greater than N?
    no, we want N to be bigger than that.

    think you can plug it into a proof now? follow my example.
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  14. #14
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    After several more hours writing out the examples i cant grasp it! The idea of chosing maths as a degree choice is fading!

    Can anyone recommend good texts to read that might shed new light for me?
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  15. #15
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    Quote Originally Posted by aceband View Post
    After several more hours writing out the examples i cant grasp it! The idea of chosing maths as a degree choice is fading!

    Can anyone recommend good texts to read that might shed new light for me?
    What level do you wish to learn at? Are you doing calculus, analysis, or what?
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