1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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 ;)

  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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook