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Sequences and Series

  1. Apr 7, 2014 #1
    1. The problem statement, all variables and given/known data


    So, I actually have a bunch of these problems and I cannot do any of them. I don't think I'm really understanding it. Here is the question: (one of them) The way I wrote them, a_n means a sub n

    For each sequence a_n find a number k such that n^k a_n
    has a finite non-zero limit.
    (This is of use, because by the limit comparison test the series ∑from 1 to infinity (a_n) and ∑from 1 to infinity (n^-k) both converge or both diverge.)

    The question:
    a_n= (6+3n)^-7
    What does k equal?

    2. Relevant equations
    N/A


    3. The attempt at a solution
    I'm actually 100% lost on this problem. I don't even know where to start :(
    Here's my attempt. It may not make any sense because I don't understand this. But here it is:
    (6+3n)^-7>0
    1/(6+3n)^7>0
    1>0
    Ans: 1
    But that is incorrect.
     
  2. jcsd
  3. Apr 7, 2014 #2

    Dick

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    Here's a hint. (6+3n)=n(6/n+3). Try and do something with that.
     
  4. Apr 7, 2014 #3
    Would you mind explaining the meaning? How would I find k from it?
     
  5. Apr 7, 2014 #4

    Dick

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    Ok, so (6+3n)^(-7)=(n(6/n+3))^(-7)=n^(-7)*(6/n+3)^(-7). What's the limit n->infinity of (6/n+3)^(-7)?
     
  6. Apr 7, 2014 #5
    isn't it infinity?
     
  7. Apr 7, 2014 #6

    Dick

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    I don't think so. Why do you? What's limit 6/n as n->infinity?
     
  8. Apr 7, 2014 #7
    because isnt it the same as saying ((n+3)/6)^7 so as n approaches infinity, it approaches infinity?
     
  9. Apr 7, 2014 #8
    but how would doing that help me find what k is?
     
  10. Apr 7, 2014 #9

    Dick

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    No, it isn't. ((6/n)+3)^(-7)=1/((6/n)+3)^7. It's not 6/(n+3). It's (6/n)+3. They are very different.
     
  11. Apr 7, 2014 #10
    and using this new limit (6/n)+3, how would it lead me to solve for k?
     
  12. Apr 7, 2014 #11

    Dick

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    I think you should stop asking that until you do the algebra correctly and tell me what limit n->infinity ((6/n)+3) is. Then take a breath and think about it.
     
  13. Apr 7, 2014 #12
    the limit is 3 because the 6/n goes to 0
     
  14. Apr 7, 2014 #13

    Dick

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    Ok, so what's limit ((6/n)+3)^(-7). Have you thought about what that might mean for your question of what k is?
     
  15. Apr 7, 2014 #14
    1/3^7= 1/2187.
    I still dont know where k comes into play.
     
  16. Apr 7, 2014 #15
    Is k 7?
     
  17. Apr 7, 2014 #16
    Where did you get the n(6/n+3) here? Why was it necessary to factor out the n?
     
  18. Apr 7, 2014 #17

    Dick

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    Ok, so (6+3n)^(-7)=(n(6/n+3))^(-7)=n^(-7)*(6/n+3)^(-7). As n->infinity (6/n+3)^(-7)->1/3^7. Does it make sense to you that k=7 works??
     
  19. Apr 7, 2014 #18

    Dick

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    Because you want the series to look like n^(-k) times something that approaches a nonzero constant. It's a good idea to factor out the n and see what's left.
     
  20. Apr 7, 2014 #19
    yes it does. so just to recap, i had to factor out an n to create the n^k? what if n were in the demonimator instead?
     
  21. Apr 7, 2014 #20

    Dick

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    Then you would factor out a 1/n^(k)=n^(-k). It just changes the sign of k. Doesn't it?
     
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