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Lim as n approaches infinity for n/2^n

  1. Jun 28, 2012 #1
    1. Determine whether the sequence converges or diverges. If it converges, find its limit.

    {[itex]\frac{n}{2n}[/itex]}+∞n = 1

    I know that we have to use L'Hopital's rule for this, because as n increases, both the numerator and the denomator approach infinity. However, doing that gives me [itex]\frac{1}{n*2n-1}[/itex], which complicates things further.

    I have a calc II quiz on Monday, so any help would be appreciated.
     
  2. jcsd
  3. Jun 28, 2012 #2

    Mark44

    Staff: Mentor

    You are not differentiating 2n correctly. You are using the power rule, but 2n is an exponential function, not a power function. It's true that d/dx(x3) = 3x2, but it is NOT true that d/dx(2x) = x*2x - 1, which is what you did.

    Rewrite 2n as en*ln(2)
     
  4. Jun 28, 2012 #3
    Okay. I know that [itex]\frac{dx}{dy}[/itex] of ex is ex.

    So for 2n, it can be rewritten as en*ln2.

    Let u = n*ln 2, so en*ln2 becomes eu.

    Differentiating this becomes eu.

    And differentiating u should...be done using the product rule correct?

    Which should give (1)(ln 2) because differentiating ln 2 becomes 0.

    This should give us a derivative, overall, of eln 2.

    Am I correct? I apologize in advance for any mistakes I might have made.
     
  5. Jun 28, 2012 #4

    Mark44

    Staff: Mentor

    Not quite. You don't take dx/dy of something (or dy/dx of something either).
    If y = ex, then dy/dx = ex.
    You need to be more specific. Differentiating with respect to what?

    d/du(eu) = eu, but to get d/dn(eu) you need to use the chain rule.

    So d/dn(eu) = eu * du/dn.
    You could, but that's more work than necessary. The constant multiple rule would be the right one. To differentiate 3x, the product rule would work, but it's simpler to do this: d/dx(3x) = 3*d/dx(x) = 3 * 1 = 3.
    No.
     
  6. Jun 28, 2012 #5
    What is the correct answer then?
     
  7. Jun 28, 2012 #6
    y = eu
    where u = n*ln2

    dy/du = eu
    du/dx = ln 2?

    Therefore
    dy/dx = ln 2*en*ln 2?
     
  8. Jun 28, 2012 #7
    Then write en*ln 2 back to the simpler form.

     
  9. Jun 28, 2012 #8
    Which becomes ln 2 * 2n.

    When you take the limit of n/2^n, using L'Hopital's rule, you get 1/ln2*2n.

    This converges to 0, because as n gets bigger, the function overall will get closer and closer to 0?

    COrrect?
     
  10. Jun 28, 2012 #9
    That's correct. You can further squeeze yourself by using ε-N to prove it:)

     
  11. Jun 28, 2012 #10
    How would we do that? We were explained the theorem in class, I didn't understand it fully however.
     
  12. Jun 28, 2012 #11

    Mark44

    Staff: Mentor

    The limit of your sequence is 0. The work you did is probably sufficient unless you were asked to use an epsilon-type argument to prove that the limit is zero.

    I'm assuming that you weren't asked to do it this way.
     
  13. Jun 28, 2012 #12
    Well, it was a part of a homework problem, but should he ask how to do it using that argument, how would that be done?
     
  14. Jun 28, 2012 #13

    Mark44

    Staff: Mentor

    Why don't you wait until he actually asks? I'm sure your instructor will show you some examples.
     
  15. Jun 29, 2012 #14
    Well, I looked at it further and found that the ε-N proof in this case would not be straightforward.

    You want [tex]\frac{n}{2^n}<\epsilon[/tex], it's not obvious how to solve the inequality. We want to reasonably magnifiy the LHS a bit to a nicer function. This may be done by invoking mean value theorem.

    [tex]\frac{2^n-1}{n}=2^{\zeta}ln2 \quad ; \quad 0<\zeta<n[/tex]. It follows that

    [tex]\frac{n}{2^n}<\frac{1}{2^{\zeta}ln2}[/tex]
    Now it looks much better.

     
  16. Jul 1, 2012 #15
    I thought about it a bit more, to do ε-N proof, I would have to accept the fact that 2^n>n^2 for n>4. I don't know how to algebraically prove it though:shy:

    With that given, the epsilon-N proof can be done by:

    We want [tex]\left|\frac{n}{2^n}\right|<\epsilon[/tex]
    Notice that
    [tex]\frac{n}{2^n}<\frac{n}{n^2}=\frac{1}{n}<\epsilon \quad for \quad n>4 [/tex]

    For any ε>0, there exists integer N=max(4,[1/ε]) such that if n>N, then [tex]\left|\frac{n}{2^n}\right|<\epsilon[/tex]
    QED
     
  17. Jul 1, 2012 #16
    Have you tried induction? You can also do 2^n≥n^2 to include n=4.
     
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