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Whether the sequence Converges

  1. Jun 26, 2012 #1
    1. Write out the first five terms of the sequence, determine whether the sequence converges, and if so, find its limit.

    {(In n)/n}+∞n = 1


    α1 = 0
    α2= 0.347
    α3= 0.366
    α4= 0.347
    α5= 0.322



    I'm not sure how to continue from this point forth. How to show whether or not the limit converges.
     
  2. jcsd
  3. Jun 26, 2012 #2
    I assume you meant {ln(n)/n}. The numerator and denominator both grow as n grows. Which one grows faster? Can you see the boundary of the sequence as n grows?
     
  4. Jun 26, 2012 #3

    LCKurtz

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    You mean whether or not the limit exists. Or, equivalently, whether the sequence converges.

    Hint: How would you do it if it was ##\frac {\ln x}x\ ?##
     
  5. Jun 27, 2012 #4
    I am honestly not sure. I know that it doesn't converge. It diverges to -∞. I just don't know how to show it. Natural logs or logs have never been my strong suit.
     
  6. Jun 27, 2012 #5
    How do you know it diverges to -∞?
     
  7. Jun 27, 2012 #6
    I was doing some reading up on this and it said to use L'Hospital's Rule. Which pretty much states that lim (n -> ∞ )[itex]\frac{f(x)}{g(x)}[/itex] = lim (n -> ∞)[itex]\frac{f'(x)}{g'(x)}[/itex].

    So, you divide both the numerator and the denominator by x (because x is the highest power in the denominator), which gives you ln(1/x)/x

    This is the same as ln 1 /x
    = ln 1 / ∞
    = 0/∞
    = 0?

    So the sequence converges to 0? This seems wrong
     
  8. Jun 27, 2012 #7
    Look up convergence tests in a book.

    Integral test, Ratio test, Comparison test.
    once you understand what the idea of these things are you will be able to do this question.
     
  9. Jun 27, 2012 #8
    Think of it as ln(n)/n=ln((n)^(1/n))... What is the limit as n^(1/n) goes to infinity? The plug that into ln
     
  10. Jun 27, 2012 #9
    I do apologize. I was looking at a similar question, {ln[itex]\frac{1}{n}[/itex]}.

    Working out this problem showed that:

    term #1 (n = 1) is 0
    term #2 (n = 2) is -0.693
    term #3 (n = 3) is -1.099
    term #4 (n = 4) is -1.386
    term #5(n = 5) is -1.609

    Taking the limit o this shows:

    lim ln[itex]\frac{1}{n}[/itex]
    = lim ln[itex]\frac{1/n}{n/n}[/itex]
    = lim ln ln[itex]\frac{1/n}{1}[/itex]
    I'm not sure how to proceed further.
     
  11. Jun 27, 2012 #10
    As for the previous question, the limit converges to 0.
     
  12. Jun 27, 2012 #11
    Those are for series convergence criteria.. Sans the comparison test
     
  13. Jun 27, 2012 #12
    We haven't been taught that yet
     
  14. Jun 27, 2012 #13
    Oh! Ln(1/n)=ln(1)-ln(n)=-ln(n)...
     
  15. Jun 27, 2012 #14
    Diverges. To - infinity. Correct? And for my original question, that converges to 0. I am sorry
     
  16. Jun 27, 2012 #15
    Since ln(n)->inf, -ln(n)->-inf
     
  17. Jun 27, 2012 #16
    Here is what I have gleamed, and understood. And I appreciate the patience you all showed while I tried to comprehend it.


    limn →∞{ln[itex]\frac{1}{n}[/itex]}

    limn →∞{ln[itex]\frac{\frac{1}{n}}{\frac{n}{n}}[/itex]

    limn →∞{ln[itex]\frac{1}{n}[/itex]}

    Which can also be written as

    limn →∞{ln[itex]\frac{1}{n}[/itex]} = ln 1 - ln n

    limn →∞{ln[itex]\frac{1}{n}[/itex]} = - ln n

    limn →∞ -ln n = -∞ as the bigger that n gets, the more negative -lnn will get.
     
  18. Jun 27, 2012 #17
    Is my answer correct?
     
  19. Jun 27, 2012 #18

    HallsofIvy

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    "So"??? You just said to use L'Hopital's rule which has nothing to do with "divide both numerator and denominator by x". The numerator is ln(x) and its derivative is 1/x. Then denominator is x and its derivative is 1. f'/g'= (1/x)/1= 1/x. That "goes to 0" (Do NOT write "1/∞"!)

    Surely you do not believe that dividing "ln(x)" by x gives ln(1)???
     
  20. Jun 27, 2012 #19

    Curious3141

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    These first two steps are completely redundant, since you just ended up where you started.

    The rest is fine.
     
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