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Homework Help: Rigorous proof of limits of sequences (3)

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data

    But I think the definition is as follows:
    Let an be a sequence of real numbers. Then an->a iff
    for ALL ε>0, there exists an integer N such that n≥N => |an - a|< ε.

    The definition says that it has to be true for ALL ε>0, but in the example above, they just let ε to be a rational number with a very specific form. To me, the proof looks incomplete. They only proved the statement for the case of ε being rational number with that very specific form, but how about the case when we're given an irrational ε, or other rational ε that cannot be expressed in that specific form?
    Is the proof correct or not?

    2. Relevant equations

    3. The attempt at a solution
    Shown above.

    Can someone explain this? Thanks for any help! :)
    Last edited: Jan 8, 2010
  2. jcsd
  3. Jan 8, 2010 #2
    Well I'm assuming k is a real number, in which case there is nothing wrong with the proof (verify this on your own). However, there is a much simpler estimate that could have been made, namely

    [tex]\frac{1}{n+1} \leq \frac{1}{N+1} < \varepsilon[/tex]

    provided that [itex]N > \frac{1}{\varepsilon}-1.[/itex]

    Out of curiosity, which text are you learning from?
  4. Jan 8, 2010 #3
    I looked at my book again and the context before the example highly suggests that k is a natural number (or 0). It says "it suffices to consider only values for ε of the form (1/2)10-k...agree with L to at least k decimals of accuracy."

    If this is the case, is the proof still correct? If so, why?

    (I understand your much simpler proof, but I just want to know whether "the proof provided above" is complete or not.)

    The book from which this example is taken is Real Analysis by Donsig.
  5. Jan 8, 2010 #4
    Ah OK, well the author is choosing not to consider all real epsilon, which is weird. I would say it doesn't matter much since if we let epsilon -> 0 we can still make our sequence arbitrarily close to 1. Anyways it seems simpler to consider all real values of epsilon (> 0) and replace [itex]\frac{1}{2}10^{-k} = \varepsilon[/itex] with [itex]\frac{1}{2}10^{-k} < \varepsilon.[/itex]
  6. Jan 9, 2010 #5


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    Given any [itex]\epsilon> 0[/itex], there exist k such that [itex](1/2)10^{-k}< \epsilon[/itex]. Then if [itex]|a_n- L|< (1/2)10^{-k}[/itex], it is less than [itex]\epsilon[/itex] so this is sufficient.

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