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Sequence e^2/2^n, converge or diverge (easy)

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    does the series [itex]e^n/2^n[/itex] converge or diverge?

    does the series [itex]2^n/e^n[/itex] converge or diverge?

    3. The attempt at a solution

    I took [itex]lim→∞ e^x/2^x[/itex] and am getting ∞, so it diverges, right?

    I also used L'Hopital's rule and got the same result.

    My prof hinted that I was wrong... what am I missing?

    Thanks!
     
    Last edited: Feb 5, 2012
  2. jcsd
  3. Feb 5, 2012 #2

    Dick

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    Yes, you are right. e/2 is greater than 1. The series diverges.
     
  4. Feb 5, 2012 #3
    what about 2/e?
     
  5. Feb 5, 2012 #4

    Dick

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    You tell me. What do you know about the geometric series r^n?
     
  6. Feb 5, 2012 #5
    8point1,

    there is a distinction. Those expressions are the general terms
    of the sequences and the series.

    The summations of those terms are the respective series.
     
  7. Feb 5, 2012 #6
    The sequence diverges, so the series must diverge also, correct?
     
  8. Feb 5, 2012 #7

    Dick

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    That's correct for (e/2)^n. (2/e)^n is different.
     
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