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HW Check: Prove that the sequence {a_n} converges to 1/2

  1. Sep 4, 2011 #1
    Hi,

    Just hoping someone could check my work and point out any errors, if any.

    1. The problem statement, all variables and given/known data

    Consider the sequence {[itex]a_n[/itex]} defined by [itex]a_n=\frac{n}{2n+\sqrt{n}}[/itex]. Prove that [itex]\lim_{x\to\infty}a_n=\frac{1}{2}[/itex]. (Do NOT use any of the "limit rules" from Section 2.2.)

    2. Relevant equations

    A sequence [itex]a_n[/itex] is said to converge to a real number A iff for each [itex]\epsilon>0[/itex] there exists a positive integer [itex]n^*[/itex] such that [itex]\lvert a_n-A\rvert<\epsilon[/itex] for all [itex]n\geq n^*[/itex].

    3. The attempt at a solution

    Proof:

    Let [itex]\epsilon>0[/itex].

    Let [itex]n^*\in\mathbb{N}[/itex] such that [itex]n^*>[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2[/itex].

    If [itex]n\geq n^*[/itex], then

    [itex]\lvert a_n-A\rvert[/itex]

    [itex]=\lvert\frac{n}{2n+\sqrt{n}}-\frac{1}{2}\rvert[/itex]

    [itex]=\lvert\frac{2n-2n-\sqrt{n}}{2(2n+\sqrt{n})}\rvert[/itex]

    [itex]=\lvert\frac{-\sqrt{n}}{2(2n+\sqrt{n})}\rvert[/itex]

    [itex]=\lvert\frac{-n}{\sqrt{n}[2(2n+\sqrt{n})]}\rvert[/itex]

    [itex]=\lvert\frac{-n}{\sqrt{n}(4n+2\sqrt{n})}\rvert[/itex]

    [itex]=\lvert\frac{-n}{4n\sqrt{n}+2n}\rvert[/itex]

    [itex]=\lvert\frac{-n}{2n(2\sqrt{n}+1)}\rvert[/itex]

    [itex]=\lvert\frac{-1}{2(2\sqrt{n}+1)}\rvert[/itex]

    [itex]=\frac{1}{2(2\sqrt{n}+1)}[/itex]

    [itex]\leq\frac{1}{2\sqrt{n}+1}[/itex]

    [itex]<\frac{1}{2\sqrt{[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2}+1}[/itex]

    [itex]=\frac{1}{2\lvert[\frac{1}{2}(\frac{1}{\epsilon}-1)]\rvert+1}[/itex]

    [itex]=\frac{1}{2[\frac{1}{2}(\frac{1}{\epsilon}-1)]+1}[/itex]

    [itex]=\frac{1}{(\frac{1}{\epsilon}-1)+1}[/itex]

    [itex]=\frac{1}{\frac{1}{\epsilon}}[/itex]

    [itex]=\epsilon[/itex].
     
  2. jcsd
  3. Sep 4, 2011 #2

    dynamicsolo

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    Homework Helper

     
    Last edited: Sep 4, 2011
  4. Sep 4, 2011 #3
    Thanks dynamicsolo.

    And, yeah, I get the substitution by knowing [itex]\frac{1}{2\sqrt{n^*}+1}<\epsilon[/itex]. Solving for epsilon, [itex][\frac{1}{2}(\frac{1}{\epsilon}-1)]^2<n^*[/itex]. And since [itex]n\geq n^*>[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2[/itex], then [itex]\frac{1}{2\sqrt{n}+1}<\frac{1}{2\sqrt{[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2}+1}[/itex].
     
  5. Sep 4, 2011 #4

    dynamicsolo

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    Homework Helper

    That's OK -- everyone does that in these epsilon-delta proofs. Your choice should be acceptable because [itex]\epsilon(n) = \frac{1}{1 + 2\sqrt{n}}[/itex] is one-to-one, so there is no ambiguity in transforming between epsilon and n , and the limit is zero as n goes to infinity.
     
  6. Sep 4, 2011 #5
    Yeah, I figured there was no problem in the substitution, though, maybe I should have provided the reasoning behind it in the proof. Oh well.

    Anyway, thanks again.
     
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