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Thread: choosing n given epsilon

  1. #1
    Junior Member
    Sep 2007

    choosing n given epsilon

    I asked this before and the guy said it was divergent; however, it's clearly not. I just wanted some help on how to do it.

    I need to find an N(epsilon) for the following sequence such that for all n>=N(epsilon) the absolute value of the nth element is less than epsilon >0:

    n(sqrt((n^4)+4) - n^2) (i.e. the n is multiplied to both the sqrt((n^4)+4)) and n^2 with subtraction between those two terms. I know that it converges to zero. I just don't know how to find a sequence (it must be strictly decreasing) that bounds it and makes finding the N(epsilon) easy.

    I appreciate any help.
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    $\displaystyle a_n=n(\sqrt{n^4+4}-n^2)=\frac{n(n^4+4-n^4)}{\sqrt{n^4+4}+n^2}=\frac{4n}{\sqrt{n^4+4}+n^2 }<\frac{4n}{n^2}=\frac{4}{n}<\epsilon$
    Then $\displaystyle n>\frac{4}{\epsilon}$.
    So, take $\displaystyle n_{\epsilon}=\left[\frac{4}{\epsilon}\right]+1$
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