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Real Analysis question

  1. Jan 21, 2012 #1
    Let n ≥ 1 be an integer and ε > 0 a real number. Without making reference or use of nth roots, prove that there exists a positive integer m such that

    [tex]\left (1- \frac{1}{m} \right )^{n}> 1-\varepsilon [/tex]

    How would I go about proving this? Would I just solve for m?
     
  2. jcsd
  3. Jan 21, 2012 #2

    tiny-tim

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    Hi BrownianMan! :smile:
    Yup! :biggrin:
     
  4. Jan 21, 2012 #3

    SammyS

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    Just solving for m would require use of an nth root, right ? ... So, "No." to that.
     
  5. Jan 21, 2012 #4

    Ray Vickson

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    What are you allowed to use? For example, using Calculus you can derive the inequality [tex] (1-x)^n > 1-nx , [/tex] for [itex] 0 < x < 1.[/itex]

    RGV
     
  6. Jan 21, 2012 #5
    Can't I say

    [tex]\left ( 1-\frac{1}{m} \right )^{n} =1-\sum_{k=1}^{n}\binom{n}{k}\left ( \frac{1}{m} \right )^{k}=1-\frac{1}{m}\left (\sum_{k=1}^{n}\binom{n}{k} \left (\frac{1}{m} \right )^{k-1} \right )[/tex]
    [tex]\geq 1-\frac{1}{m}\left (2^{n}-1 \right )[/tex]

    So then

    [tex]1-\frac{1}{m}\left (2^{n}-1 \right ) > 1 - \varepsilon [/tex]
    [tex]\frac{2^{n}-1}{\varepsilon }<m [/tex]
     
  7. Jan 22, 2012 #6
    ^Would this be right?
     
  8. Jan 22, 2012 #7

    tiny-tim

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    Hi BrownianMan! :smile:
    (Your ∑ needs a (-1)k inside it. :wink:)

    Perhaps I'm missing the obvious, but where does your 2n come from? :confused:
     
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