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Epsilon-N Proof of Sequence

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that the sequence [tex](\frac{1}{1+n+n^4})[/tex] converges to 0.


    2. Relevant equations


    3. The attempt at a solution

    Given [tex]\epsilon >0[/tex], we can find [tex]n \geq N[/tex] such that:

    [tex]| \frac{1}{1+n+n^4} -0 | = \frac{1}{1+n+n^4} < \frac{1}{1+n}< \epsilon[/tex]

    Now what value of N should we take to complete the proof? And why?

    This is what I guess:
    We have [tex]\frac{1}{1+n}< \epsilon[/tex] so,

    [tex]n+1> \frac{1}{\epsilon}[/tex]

    [tex]n>\frac{1}{\epsilon} -1[/tex]

    [tex]N = \frac{1}{\epsilon} -1[/tex]

    Is this right? I appreciate it if anyone could provide me with some explaination.
     
  2. jcsd
  3. Oct 28, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    i think that will work, in that for the N you give you know every term with n>N will be less than epsilon.... however, i think you can put much tighter constraints on N using the n^4 part of the sequence.

    1/n^4 goes to zero a lot quicker than 1/n. And for n>>1, n^4 + n + 1 looks more like n^4 than n
     
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