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Math Help - Infinite sequence help

  1. #1
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    Infinite sequence help

    Hi.

    I have been given this sequence:



    I have to find if it converges or diverges as n -> infinity

    What i did:

    Divided numerator/denominator by root n.

    [sin (6n)/root n]/[(8/root n) + 1]

    as n -> infinity, the bottom converges to 1.

    I'm not sure how to evaluate the top. How do you evaluate sin infinity? sin can vary between -1 to 1.
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  2. #2
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    Quote Originally Posted by Kuma View Post
    Hi.

    I have been given this sequence:



    I have to find if it converges or diverges as n -> infinity

    What i did:

    Divided numerator/denominator by root n.

    [sin (6n)/root n]/[(8/root n) + 1]

    as n -> infinity, the bottom converges to 1.

    I'm not sure how to evaluate the top. How do you evaluate sin infinity? sin can vary between -1 to 1.
    The main point of the sine function is that it does not become infinite. Left on its own \displaystyle \lim_{n \to \infty} sin(6n) would not converge since it never settles to a specific value. On the other hand

    \displaystyle \frac{sin(6n)}{8 + \sqrt{n}} \approx \frac{sin(6n)}{\sqrt{n}}
    for large n. n approaches infinity, making the denominator get very large, while sin(6n) varies between 1 and -1. Thus the limit goes to 0.

    -Dan
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  3. #3
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    Hi. Thanks for the help!
    Another sequence question I'm stuck on



    Given that, how would you evaluate the sequence given n approaches infinity.
    I thought it would converge to 0 but that's wrong. I tried dividing the top and bottom by n!, resulting in 3/n! on the top and 2^n/n! on the bottom.

    It seems like an indeterminate form just looking at it. The top would approach infinity and so would the bottom. So using L'hopitals rule, would deriving the top give me 0? It looks like a constant for any n; and deriving the bottom would give me:
    n(2^n-1). Which also results in a convergence of 0.
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  4. #4
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    Quote Originally Posted by Kuma View Post
    Hi. Thanks for the help!
    Another sequence question I'm stuck on
    Here is a hint: If n\ge 6 then 2^n<(n-1)!.
    Note that 3\cdot n!=3\cdot n\cdot (n-1)!.
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