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Converge or Diverge

  1. May 5, 2010 #1
    1. The problem statement, all variables and given/known data

    does the series (Sigma) 4^n/e^n^5 converge or diverge?

    2. Relevant equations



    3. The attempt at a solution[/b

    just got out of a test where this was asked and my teacher told me I had to use the integral test however I think using direct comparison to a known geometric series could work as well. here is what i thought of...

    4^n/e^n^5 < 1/e^n^5 , where the latter is a convergent geometric series (constant raised to a variable power) that n^5 is throwing me off but does my reasoning make sense at all?
     
  2. jcsd
  3. May 5, 2010 #2

    Mark44

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    Your inequality is going the wrong way. 4^n > 1 for n > 1, so 4^n/e^n^5 > 1/e^n^5 for n > 1. Now you have the situation where the arbitrary term of your series is larger than the corresponding term of a convergent series, so this comparison doesn't do you any good.
     
  4. May 5, 2010 #3

    LCKurtz

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    If your teacher suggests the integral test I would bet that the problem is mis-printed and should have been

    [tex]\sum \frac{n^4}{e^{n^5}}[/tex]
     
  5. May 5, 2010 #4
    Yep, the integral test would be best here. Do a u-substitution where u = x^5 and you can go from there
     
  6. May 6, 2010 #5
    Is there any other way to do it other than the integral test?
     
  7. May 6, 2010 #6

    LCKurtz

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    Have you asked your teacher if the problem was mis-printed as I suspect?
     
  8. May 6, 2010 #7
    Kurtz u r right I copied it onto this forum incorrectly.

    On another note, is 1/e^n^5 a geo series?
     
  9. May 6, 2010 #8

    LCKurtz

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    That is ambiguous as written. I suppose you mean:

    [tex]\frac 1 {e^{(n^5)}}[/tex]

    Remember that a sequence {an} is geometric if

    [tex]\frac{a_{n+1}}{a_n} = r[/tex]

    a common ratio r, for all n. Does that work for your sequence?
     
  10. May 6, 2010 #9
    I honestly don't know, tho I suppose u could show convergence by saying 1/e^(n^5) < 1/e^n ? Where l 1/e l < 1 ?

    On another note, if u used the root test on the series we were talking about would u get:

    n^(4/n) / e^(1^5) ((divide all n's by n))

    which as n -> infinity the expression goes to 1/e?

    That is what I originally tried on my test and she told me to start over after looking at it. :/
     
  11. May 6, 2010 #10

    LCKurtz

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    If you read the rules of these forums you will see they discourage "baby talk". It doesn't take much more energy to type the whole words, which you mostly do.
    What about the n4 in the numerator?

    No, the denominator is wrong.
     
  12. May 6, 2010 #11
    Sorry my computer won start so I'm typing on an iPhone.

    Thanks for the help. I guess I'll just have to deal with the integral test.
     
  13. May 6, 2010 #12

    LCKurtz

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    Not much to deal with; the problem is all set up for it and it's easy.

    :wink:
     
  14. May 6, 2010 #13
    True. I just like trying to tackle things in more ways than one. I find it expands my understanding of the concepts better.

    Thanks again for you responses. Much appreciated! :)
     
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