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Basic problem with supremum question.

  1. Dec 14, 2011 #1
    Hey guys!

    Revising for an exam and I've come across a pretty basic problem.

    Question: Prove that the supremum of the set A : { 3n / (5n+1) :n€N} is 3/5

    My answer: So 3n / (5n+1) ≤ 3n / 5n = 3/5 so 3/5 is an upper bound.

    Now, We claim that 3/5 is the least upper bound. Take β < 3/5 so now I need a positive integer n > .........This is bit I don't know how to do.... (how do I choose this part?)
    I then know that you re-arrange n>........ to see that the β we chose earlier is: β < 3n / (5n+1) which is impossible, hence Sup(A) = 3/5

    I hope you understand what I mean..
    Regards as always
    Tam
     
  2. jcsd
  3. Dec 14, 2011 #2

    radou

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    Homework Helper

    You should rearrange your inequality somehow to show that for any given ε > 0, you can find some natural number N such that 3N / (5N + 1) > 3/5 - ε. I haven't tried it out myself, but this should be the concept, unless I'm mistaken.
     
  4. Dec 15, 2011 #3
    Are you sure? The solutions I have are vague to say the least, but would you have any idea how I could do this?
     
  5. Dec 15, 2011 #4

    HallsofIvy

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    Staff Emeritus
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    Solve the inequality
    [tex]\frac{3N}{5N+1}> \beta[/tex]
    for [itex]\beta< 3/5[/itex]
    or, equivalently,
    [tex]\frac{3N}{5N+1}> \frac{3}{5}- \epsilon[/tex]
    for N.
     
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