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Convergence of a Sequence

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Be [itex]K \geq 1[/itex]. Conclude out of the statement that [itex]\lim_{n \to \infty } [/itex] [itex]\sqrt[n]{n} = 1[/itex], dass [itex]\sqrt[n]{K} = 1[/itex]


    3. The attempt at a solution
    [itex]\lim_{n \to \infty } \sqrt[n]{K} \Rightarrow 1 \leq \sqrt[n]{K} \geq 1 + ...[/itex]

    I got issues with the right inequality, where the 3 dots are. I´m not sure if just insert the [itex] \sqrt[n]{n} [/itex] there and that s about it.

    Thanks in advance ;)

    Christian....
     
  2. jcsd
  3. Nov 4, 2012 #2
    Let [itex]1 \leq K \leq n[/itex] for some big n (we're going to let it tend to infinity later)
    Then we can do a squeeze:
    [tex]\lim_{n\to\infty}\sqrt[n]{1} \leq \lim_{n\to\infty}\sqrt[n]{K} \leq \lim_{n\to\infty}\sqrt[n]{n}[/tex]
     
  4. Nov 4, 2012 #3
    Many thanks Fightfish for your quick reply so i can state the following:

    For the 3 real Sequences [itex]\sqrt[n]{1} , \sqrt[n]{K} , \sqrt[n]{n}[/itex] [itex]\exists N [/itex] [itex] \forall n \geq N[/itex] is [itex]\sqrt[n]{1} \leq \sqrt[n]{K} \leq \sqrt[n]{n}[/itex]

    and [itex]\lim_{n\to\infty}\sqrt[n]{1} = \lim_{n\to\infty}\sqrt[n]{n}[/itex].


    [itex]\Rightarrow \sqrt[n]{K} [/itex] converges and [itex]\lim_{n\to\infty}\sqrt[n]{K} = \lim_{n\to\infty}\sqrt[n]{n}[/itex]
     
  5. Nov 5, 2012 #4
    I thought about this and maybe this one here is more elegant than the other one, would be cool if someone could backcheck it.

    [itex] \forall n \text{ with } n > K >1: \sqrt[n]{K} <\sqrt[n]{n}[/itex]

    and [itex]\lim_{n\to\infty}\sqrt[n]{1}= 1[/itex]

    [itex] \lim_{n\to\infty}\sqrt[n]{K} \leq \lim_{n\to\infty}\sqrt[n]{n} = 1 [/itex]

    Thanks again for your helping...
     
  6. Nov 5, 2012 #5

    SammyS

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    Hello Christian (Dodobird). Welcome to PF !
    I also have issues with the inequality:

    [itex]\displaystyle \lim_{n \to \infty } \sqrt[n]{K}\ \Rightarrow \ 1 \leq \sqrt[n]{K} \geq 1 + ...[/itex]

    Why do you have both ≤ and ≥ in the same compound inequality?
     
  7. Nov 5, 2012 #6
    Thank you Sammy for your warm welcome
    Oh yeah, you are right. Both signs should point in the same direction. I mistakenly wrote it in the wrong way.Sorry about that.
    So it should be:

    [itex]\displaystyle \lim_{n \to \infty } \sqrt[n]{K}\ \Rightarrow \ 1 \leq \sqrt[n]{K} \leq 1 + ...[/itex]


    Thx ;)
     
  8. Nov 5, 2012 #7

    SammyS

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    I'm just making sure that I understand this exercise.

    You are to prove, for K≥1 that [itex]\lim_{n \to \infty }\sqrt[n]{K} = 1\,,[/itex] using the result that [itex]\lim_{n \to \infty }\sqrt[n]{n}=1\ .[/itex] Is that correct?
     
  9. Nov 5, 2012 #8
    Yeah, that´s correct Sammy. Do you see any flaws?
    I´m pretty new to proofs in Mathematics and still struggle with it and still feel a little bit insecure when I got to prove something.
     
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