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Math Help - Sequence

  1. #1
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    Sequence

    I can't do this one :-(

    Prove from the definition that  a_n=\frac{n^2 + 3n}{n^2 +2} -> 1 as n -> \infty

    My lecturer wrote on the board that we should be answering the question as follows (he never gave us an example though!):

    For a given positive h you must ensure that the difference of and 1 can be made to stay less than h. When you have an expression for the difference, look for a related simpler expression that can easily be made less than h.

    Haven't got a clue, please help.
    Last edited by Natasha1; May 23rd 2006 at 08:35 AM.
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  2. #2
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    Quote Originally Posted by Natasha1
    I can't do this one :-(

    Prove from the definition that  a_n=\frac{n^2 + 3n}{n^2 +2} -> 1 as n -> \infty
    My lecturer wrote on the board that we should be answering the question as follows (he never gave us an example though!):
    For a given positive h you must ensure that the difference of and 1 can be made to stay less than h. When you have an expression for the difference, look for a related simpler expression that can easily be made less than h.
    Haven't got a clue, please help.
    Hello,

    I haven't much time now, so I show you the steps without much explanation:
    Calculate the difference:
      \frac{n^2 + 3n}{n^2 +2} -1 \leq h Use synthetic division.

      1+\frac{3n -2}{n^2 +2} -1 \leq h

      \frac{n(3 -{2\over n})}{n(n +{2\over n})}\leq h

      \frac{3 }{n}\leq h

    Greetings

    EB
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  3. #3
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    Quote Originally Posted by Natasha1
    I can't do this one :-(

    Prove from the definition that  a_n=\frac{n^2 + 3n}{n^2 +2} -> 1 as n -> \infty

    My lecturer wrote on the board that we should be answering the question as follows (he never gave us an example though!):

    For a given positive h you must ensure that the difference of and 1 can be made to stay less than h. When you have an expression for the difference, look for a related simpler expression that can easily be made less than h.

    Haven't got a clue, please help.
    Definition? What should be the definition here? I am lost.

    Suppose we just show that a_n tends to 1 as n approaches infinity, without using that "definition"?

    How about this way.
    a_n = (n^2 +3n) / (n^2 +2) --> 1, as n-->infinity.
    Divide both numerator and denominator by n^2,
    a_n = [(n^2)/(n^2) +(3n)/(n^2) ] / [(n^2)/(n^2) +2/(n^2)] --->1, as n-->infinity.
    a_n = [1 +3/n] / [1 +2/(n^2)] --> 1, as n-->infinity.
    The 3/n and 2/(n^2) approach zero as n approaches infinity, so,
    a_n = [1] / [1] ---> 1
    a_n --> 1

    Shown.
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  4. #4
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    Quote Originally Posted by ticbol
    Definition? What should be the definition here? ...
    Hello,

    I guessed that Natasha1 refered to Cauchy's convergence criteria.

    May be I was wrong, but finally I got the simplier expression which was asked for.

    Greetings

    EB
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