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Sequence diverges or converges

  1. May 4, 2014 #1
    Question 1

    write the first 4 terms in the sequence defined by a 1 = -2, a n+1 = an/n!

    this what I tried

    a2 = -2/1
    a3 = -2/2
    a4 = -1/6

    Question 2

    determine the following sequences converges or diverges

    1- an = sqrt(n^2 -3 )/ 5th rt(n^2)

    what I would try is to divide both top and bottom by n^2 but here doesn't work that
    way because the denominator is 5th rt

    2- a1 = 3, an+1 = (-1)^n+1 an + 3

    a2 = 6
    a3 = -3

    but the ans would be ( 3,0,3,0,3,0)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 4, 2014 #2

    jbunniii

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    Looks fine to me.
    When ##n## is large, ##\sqrt{n^2 - 3}## is nearly the same as ##\sqrt{n^2}##. So you can use ##\sqrt{n^2}## instead of ##\sqrt{n^2 - 3}## to determine whether the sequence converges or not. Then you can make the argument more precise by comparing ##\sqrt{n^2 - 3}## with an appropriate sequence. (Hint: try comparing it with ##\sqrt{n^2}/2##.)
    OK, so what can you conclude? Does the sequence converge?
     
  4. May 4, 2014 #3
    question 2

    A-
    actually this section was before the comparison theorem so let's assume we can't compare it for now

    what is the other way


    B-it diverges but I don't know how to come up with these values as my values differ ( 3,0,3,0,3,0)
     
  5. May 4, 2014 #4

    jbunniii

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    OK, try simplifying: ##\sqrt{n^2 - 3}/\sqrt[5]{n^2} = (n^2 - 3)^{1/2}n^{-2/5} = (n^2 - 3)^{1/2}(n^{-4/5})^{1/2} = ?##

    Check your values of ##(-1)^{n+1}##. This is ##-1## when ##n## is even, and ##+1## when ##n## is odd. I suspect you have it reversed.
     
  6. May 4, 2014 #5

    for part A, now it looks complicated to me than before what my prof did is

    moving the bottom to the numerator and you get infinity

    for part B I still unable to get ( 3,0,3,0,3,0)
     
  7. May 4, 2014 #6

    jbunniii

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    Can you show what you mean by this?
    Let's go back to the definition:
    To get ##a_2## we plug in ##n=1##: this gives us ##a_2 = (-1)^2 a_1 + 3 = (1)(3) + 3 = 6##. So already this is different from your ##3,0,3,0,...##.
     
  8. May 4, 2014 #7
    ( 3,0,3,0,3,0) is my professor answer to this part

    but as you see we are not getting close to that
     
  9. May 4, 2014 #8

    LCKurtz

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    I'm guessing there is a typo somewhere and the recursion is supposed to be$$
    a_1 = 3, a_{n+1} = (-1)^{n} a_n + 3$$
     
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