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Can't find the expression for comparision test on a serie

  1. Jan 11, 2013 #1
    Hi guys

    I'm trying to find out the nature of this serie

    [tex]\sum_{n=1}^{\infty}\left[ e^{1/n}-e^{-1/n}\right][/tex]

    I know the ratio and root tests are inconclusive, just the comparision test works (it diverges)

    But with which expression shall I compare?

    Thank you so much
     
  2. jcsd
  3. Jan 11, 2013 #2

    tiny-tim

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    hi joao_pimentel! :smile:

    have you tried ∑ 2/n ?

    (yes, i know it doesn't work, but why doesn't it work, and how might you fix it? :wink:)
     
  4. Jan 11, 2013 #3
    Hi

    Thanks for the reply :)

    I'm trying to find this limit

    [tex]\lim\frac{e^{1/n}-e^{-1/n}}{\frac{2}{n}}[/tex]

    let's see what can I get...
     
  5. Jan 11, 2013 #4
    ok, the limit is zero

    [tex]\lim\frac{e^{1/n}-e^{-1/n}}{\frac{2}{n}}=\frac{e^{0}-e^{0}}{\frac{2}{0}}=0[/tex]

    humm, what changes shall I make?!?
     
    Last edited: Jan 11, 2013
  6. Jan 11, 2013 #5

    tiny-tim

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    no, i mean is e1/n - e-1/n greater or less than 2/n ?

    and by roughly how much? :smile:
     
  7. Jan 11, 2013 #6
    it doesn't work because 2/n>e^(1/n)-e^(-1/n)

    and how can I fix it? :)
     
  8. Jan 11, 2013 #7

    tiny-tim

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    that's right, it's less :smile:

    (and you need it to be greater than something, to prove it diverges)

    but roughly how much less is it? :wink:
     
  9. Jan 11, 2013 #8
    humm, intresting, I'm realising that both expression are really close as n goes inf.

    ok, my last limit was incorrect, sorry

    [tex]\lim\frac{e^{1/n}-e^{-1/n}}{\frac{2}{n}}=\frac{e^{0}-e^{0}}{\frac{2}{\infty}}=\frac{1-1}{0}=\frac{0}{0}=Ind[/tex]

    and solving it it gives 1
     
  10. Jan 11, 2013 #9
    Thank you tiny-tim

    of course, I may compare it with 1/x :)
     
  11. Jan 12, 2013 #10
    By the way, how did you find out that 2/n and the other expression are very similar as n goes to inf?

    By the Mc-Lauren expansion serie?

    Thank you
     
  12. Jan 12, 2013 #11

    tiny-tim

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    yes of course :smile:
     
  13. Jan 13, 2013 #12
    Thank you so very much again :)

    you've been most helpful
     
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