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Epislon Delta Proof: Limit of a Sequence

  1. May 11, 2007 #1
    1. The problem statement, all variables and given/known data

    prove: lim n-> inifity Sn = 0

    Let Sn = (n+1)/(n^2 +1)

    2. Relevant equations

    (for all epsilon > o) (there exists N) (for all n) [ n>N => |Sn-0| < epsilon]


    3. The attempt at a solution

    i let epsilon be arbitrary, so we must show that there exists an N such that for all n [ n>N => |Sn-0| < epsilon

    Find N

    |Sn -0| = |(n+1)/(n^2 +1)| = (n+1)/(n^2 +1)

    I'm completely stuck on this step. I'm not sure how to deal with inequalities where i can make it (some term / n) < epsilon

    so I can't choose N= term/epsilon

    Any help would be much appreciated
     
  2. jcsd
  3. May 11, 2007 #2

    NateTG

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    Hint:
    [tex]\frac{n+1}{n^2+1} < \frac{n+1}{n^2-1}[/tex]
     
  4. May 11, 2007 #3
    with that in mind

    do i say

    n>2

    so that 1. n^2 - 1 > 0
    2. n+1 < n+n = 2n

    n^2 >4
    1/4 n^2> 1
    -(1/4) n^2 < -1
    n^2 -1> n^2 - (1/4) n^2 = (3/4) n^2

    1/(n^2-1) < 4/(3n^2)

    |Sn|< (2n)/(n^2-1) < (8n)/(6n^2)=4/(3n)

    4/(3n) < epsilon
    when
    n> 4/(3 epsilon)

    N = max {2, 4/(3 epsilon)}
     
  5. May 11, 2007 #4

    NateTG

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    What you have looks ok.

    I was actually thinking:
    [tex]S_n=\frac{n+1}{n^2+1}<\frac{n+1}{n^2-1}=\frac{n+1}{(n+1)(n-1)}=\frac{1}{n-1}[/tex]
    Which leads to
    [tex]N=\max(2,\frac{1}{\epsilon}+1)[/tex]

    Another good one is:
    [tex]S_n=\frac{n+1}{n^2+1}\leq\frac{2n}{n^2}=\frac{2}{n}[/tex]
    (The inequality is clearly true since the numerator increases, and the denominator decreases, and both are positive.)
    [tex]N=\max(1,\frac{2}{\epsilon})[/tex]
     
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