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Thread: [SOLVED] Epsilon-N Proof

  1. #1
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    [SOLVED] Epsilon-N Proof

    Prove that the sequence $\displaystyle (\frac{1}{1+n+n^4})$ converges to 0.

    Here's the solution:



    Here's my problem: I don't understand why they wrote $\displaystyle N=\frac{1}{\epsilon}$ !

    We know that $\displaystyle n+1 > \frac{1}{\epsilon}$

    Therefore $\displaystyle n> \frac{1}{\epsilon} -1$

    And so we should take: $\displaystyle N= \frac{1}{\epsilon} -1$

    Isn't that right???

    Also, I don't understand why they first omitted $\displaystyle n^4$ from the denominator when they could omit $\displaystyle n+1$ instead. Because I think $\displaystyle \frac{1}{n^4}$ goes to zero a lot quicker. I appreciate some explanation.
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  2. #2
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    Quote Originally Posted by Roam View Post
    We know that $\displaystyle n+1 > \frac{1}{\epsilon}$

    Therefore $\displaystyle n> \frac{1}{\epsilon} -1$

    And so we should take: $\displaystyle N= \frac{1}{\epsilon} -1$
    No. Firstly, $\displaystyle N$ must be an integer. $\displaystyle \frac1\epsilon-1$ may not be an integer. Secondly, $\displaystyle n+1>\frac1\epsilon$ is what we want, not what we know. By choosing $\displaystyle N=\left\lceil\frac1\epsilon\right\rceil,$ we can be assured that whenever $\displaystyle n>N,$ we have $\displaystyle n+1>\frac1\epsilon.$
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  3. #3
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    Quote Originally Posted by proscientia View Post
    $\displaystyle \frac1\epsilon-1$ may not be an integer.
    Why is that?

    And, what about my second question? Why couldn't we start out by omitting $\displaystyle 1+n$ instead of $\displaystyle n^4$?

    $\displaystyle \frac{1}{1+n+n^4} < \frac{1}{n^4} < \epsilon$

    Thus $\displaystyle N= \frac{1}{\epsilon ^4}$
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  4. #4
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    Quote Originally Posted by Roam View Post
    Thus $\displaystyle N= \frac{1}{\epsilon ^4}$
    That does not work.
    There is no reason that $\displaystyle \frac{1}{\epsilon ^4}$ is an integer.
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  5. #5
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    Quote Originally Posted by Plato View Post
    That does not work.
    There is no reason that $\displaystyle \frac{1}{\epsilon ^4}$ is an integer.
    Then how do you know that $\displaystyle \frac{1}{\epsilon}$ is an integer?
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  6. #6
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    Where is it stated that $\displaystyle \frac1\epsilon$ is an integer? $\displaystyle N$ is $\displaystyle \left\lceil\frac1\epsilon\right\rceil,$ not $\displaystyle \frac1\epsilon.$
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  7. #7
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    Quote Originally Posted by Roam View Post
    Then how do you know that $\displaystyle \frac{1}{\epsilon}$ is an integer?
    You don't! If, for example, $\displaystyle \epsilon= .00000003= \frac{3}{100000000}$ then $\displaystyle \frac{1}{\epsilon}= \frac{100000000}{3}$ which is not an integer.

    If you look in any text book you should see that they say something like $\displaystyle N> \frac{1}{\epsilon} -1$, not $\displaystyle N= \frac{1}{\epsilon} -1$
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