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Math Help - Sequence problem

  1. #1
    Member GAdams's Avatar
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    Sequence problem

    Hello everyone

    I need to decide if some sequences converge or not and say what the limit is with a justification. If someone could go through the method with the first one:

    a = (6n - 3n^3) / (2n^3 + 4n^2)
    n

    thank you.
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by GAdams View Post
    Hello everyone

    I need to decide if some sequences converge or not and say what the limit is with a justification. If someone could go through the method with the first one:

    a = (6n - 3n^3) / (2n^3 + 4n^2)
    n

    thank you.
    That sequence is always less than b_n = \frac{1}{n^2} which is a p-Series convergence. Therefore, using the direct comparison test, sequence a_n converges.
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  3. #3
    Member GAdams's Avatar
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    Quote Originally Posted by colby2152 View Post
    That sequence is always less than b_n = \frac{1}{n^2} which is a p-Series convergence. Therefore, using the direct comparison test, sequence a_n converges.
    I don't understand what you wrote.

    How do you arrive at:

    sequence is always less than b_n = \frac{1}{n^2} ?
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  4. #4
    GAMMA Mathematics
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    Actually, the comparison test will not work here because the sequence you gave is negative. You could treat the n's as x's and split the fraction into two terms. Integrate both terms separately. Both will be positive decreasing functions, so the integral test passes, and you will have two convergence sequences added (one is negative) to each other.

    Also, keep in mind that the limit of this function is -1.5, so it definitely converges!
    Last edited by colby2152; January 2nd 2008 at 01:10 PM.
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  5. #5
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    Quote Originally Posted by colby2152 View Post
    The sequence is equal to one half at n=1. At values of n greater than one, the sequence is negative, so therefore it is always less than the sequence I compared it to.
    Does that hold generally?
    If i get a sequence whose every term is less than the corresponding term of a convergent sequence, is the sequence convergent?

    consider (-1,-2,-3,-4,-5,-6,.....). This sequence satisfies above conditions for the sequence \frac1{n^2} like the previous example, but is not convergent! I think we need more than that to conclude anything.

    I need to decide if some sequences converge or not and say what the limit is with a justification. If someone could go through the method with the first one:

    a = (6n - 3n^3) / (2n^3 + 4n^2)
    n
    This is just evaluating limits, isnt it?
    Doesn't the sequence converge to -\frac32 ??
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  6. #6
    GAMMA Mathematics
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    Iso, see my last reply in regards to the comparison and integral tests.
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  7. #7
    Member GAdams's Avatar
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    [QUOTE


    This is just evaluating limits, isnt it?
    Doesn't the sequence converge to -\frac32 ??[/QUOTE]


    This is what I (think) I know:
    To find out if it is convergent the limit has to be found.

    So, how do I find the limit? Is there a formula? How do you arrve at the figure -3/2?

    Thanks
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  8. #8
    Lord of certain Rings
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    I assumed you know it,

    No formula, but do you know to evaluate limits? You can also solve by sandwiching it between two sequences like -\frac32 and \frac{-3n^2}{2n^2 + n}

    Or you could just evaluate \lim_{n \to \infty} \frac{6- 3n^2}{2n^2 + n}

    I will give a explicit hint:
    \lim_{n \to \infty} \frac{\frac6{n^2}- 3}{2 + \frac1{n}}
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  9. #9
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    Quote Originally Posted by colby2152 View Post
    That sequence is always less than b_n = \frac{1}{n^2} which is a p-Series convergence. Therefore, using the direct comparison test, sequence a_n converges.
    No it is not! If you are working with series instead of sequences then \sum a_n diverges, just use the divergence theorem.
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    No it is not! If you are working with series instead of sequences then \sum a_n diverges, just use the divergence theorem.
    Yeah TPH, see my last reply to this thread. I answered it too quickly, and I may have been thinking too much since the problem was fairly simple.
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