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Show that ##\lim_{n->\infty} \frac{n^2}{2^n} = 0 ##

  1. Feb 28, 2017 #1
    1. The problem statement, all variables and given/known data
    show that
    [tex]\lim_{n->\infty} \frac{n^2}{2^n} = 0 [/tex]

    2. Relevant equations

    squeeze theorem

    3. The attempt at a solution



    I tried to use squeez theorem. I dont know how to do it because dont know how to reduce [itex]2^n[/itex]

    However, I can solve this question like this.

    Given [itex]\epsilon>0[/itex], find [itex]M \in N[/itex] such that [itex] M > max (4, \frac{1}{\epsilon}) [/itex] [itex]\mid \frac{n^2}{2^n} \mid = \frac{n^2}{2^n} < \frac{n}{n^2} = \frac{1}{n} < \frac{1}{M} <\epsilon[/itex]

    [itex]2^n > n^2[/itex] and [itex]if x>4 [/itex]


    My question is how can I solve this problem with squeez theorem ?

     
    Last edited by a moderator: Mar 1, 2017
  2. jcsd
  3. Feb 28, 2017 #2

    fresh_42

    Staff: Mentor

    What about ##2^n > n^3## ?
    And I think you also need it in your prove. How became ##n^2 < n## in the above?
     
  4. Feb 28, 2017 #3
    oh.. I made mistake. but 2^n < n^3 isn't it ?
     
  5. Feb 28, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, although you then need a higher lower bound for ##M##. ##4## won't do anymore, but this isn't a problem.
    And with that, your proof works and you can also use the same estimations for the squeeze theorem.
     
  6. Mar 1, 2017 #5

    Mark44

    Staff: Mentor

    Is there some reason you need to use the squeeze theorem?

    The expression you're taking the limit of can be written as ##\left(\frac{n^{2/n}} 2\right)^n##
    The problem then boils down to showing that ##\frac{n^{2/n}} 2 < 1##, which can be done using induction.
     
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