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Math Help - finding an epsilon

  1. #1
    Junior Member
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    finding an epsilon

    I was wondering if someone could show me how to do this.

    I need to find an N(epsilon) for the following sequences such that for all n>=N(epsilon) the absolute value of the nth element is less than epsilon >0:
    1.) 1/sqrt(n)

    I think I know how to do this. For e>0, 1/sqrt(n) > e. Thus, 1>sqrt(n)*e and (1/e) < sqrt(n). Square both sides and (1/e^2) < n. I believe this is an N(e) such that n >= N(e). Is this right?

    2.) (1 + sqrt(n)) / (n^3)

    I want to say that ((1 + sqrt(n)) / (n^3)) >= (1/n^3) > e. And then this leads to n > (1/e^(1/3)). Is this right?

    3.) (sin(n)) / (2 + n^(5/3))
    4.) (sqrt((n^4) +4) - n^2) * n

    I'm not sure what to do on these. Any suggestions would be helpful.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by BrainMan View Post
    I was wondering if someone could show me how to do this.

    I need to find an N(epsilon) for the following sequences such that for all n>=N(epsilon) the absolute value of the nth element is less than epsilon >0:
    1.) 1/sqrt(n)

    I think I know how to do this. For e>0, 1/sqrt(n) > e. Thus, 1>sqrt(n)*e and (1/e) < sqrt(n). Square both sides and (1/e^2) < n. I believe this is an N(e) such that n >= N(e). Is this right?
    Because \frac{1}{\sqrt{n}} is a strictly decreasing sequence for n \ge 1,
    for any \varepsilon >0 if for some N_{\varepsilon} \in \mathbb{N}:\ \ \frac{1}{\sqrt{N_{\varepsilon}}} \le \varepsilon then for all n>N_{\varepsilon}:\ \ \ \frac{1}{\sqrt{n}} \le \varepsilon.

    So we just need to find some N \in \mathbb{N} such that:

    <br />
\frac{1}{\sqrt{N}} \le \varepsilon<br />

    So we solve:

    <br />
\frac{1}{\sqrt{x}} = \varepsilon<br />

    and take any N>x for N_{\varepsilon}, to make this specific we take:

    N_{\varepsilon}=\lceil x \rceil.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by BrainMan View Post
    2.) (1 + sqrt(n)) / (n^3)

    I want to say that ((1 + sqrt(n)) / (n^3)) >= (1/n^3) > e. And then this leads to n > (1/e^(1/3)). Is this right?

    3.) (sin(n)) / (2 + n^(5/3))
    4.) (sqrt((n^4) +4) - n^2) * n

    I'm not sure what to do on these. Any suggestions would be helpful.
    With all of these try finding a strictly decreasing sequence that bounds the
    given sequence, then apply the method used for part one to the bounding
    sequence the N_{\varepsilon} you find for the bounding sequence will then suffice
    for the given sequence.

    By the way (4) is divergent so you will not be able to find such an N_{\varepsilon}.

    RonL
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