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Deciding if the series converges or diverges

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi everyone ,, got the following question

    Sum(k=1,infinity) (2009+5^(-k^3))/(k^2(1+e^(-k^2)))


    2. Relevant equations



    3. The attempt at a solution

    first I think it diverges so I did a comparison which is :
    ak = (2009+5^(-k^3))/(k^2(1+e^(-k^2)))
    e^(k^2))/(5^(k^3)*k^2) which is less than ak ,,

    but now what should I do ?? which test should I use to make it diverges (that is if it diverges)

    note : I tried already ratio test and it didn't work and k-th term but I get (infinity over infinity) so it's no use ,, can someone help please
     
  2. jcsd
  3. Jan 2, 2010 #2

    HallsofIvy

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    It should be easy to see that
    [tex]\frac{2009+ 5^{-k^3}}{1+ e^{-k^2}}< 2010[/tex]
    compare this series to [itex]\sum 2010/k^2[/itex].

    Thanks, jgens, for catching the typo.
     
    Last edited: Jan 3, 2010
  4. Jan 2, 2010 #3
    lol ,, first of all ,, can you explain it a little further ,, how can you choose a number where the function is less than it ( i mean 20010) how did you know that number ??

    and why did you remove 5^(-k^3) and e^(-k^2) ?? do they cancel each other or what ?
     
  5. Jan 2, 2010 #4

    jgens

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    Gold Member

    Halls mistyped and probably meant that

    [tex]\frac{2009 + 5^{-k^3}}{1 + e^{-k^2}} < 2010[/tex]

    This inequality should be obvious since [itex]5^{-k^3} < 1[/itex] which implies that [itex]2009 + 5^{-k^3} < 2010[/itex]. Now, because [itex]1 + e^{-k^2} > 1[/itex] we know that

    [tex]\frac{2009 + 5^{-k^3}}{1 + e^{-k^2}} < 2010[/tex]

    Now, use the comparrison test as Halls suggested.
     
  6. Jan 3, 2010 #5
    aha ,, now I got you ... thanks :)
     
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