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Math Help - sequence problem (more of a limits question)

  1. #1
    Junior Member pirateboy's Avatar
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    [Solved] sequence problem (more of a limits question)

    Here' we're asked to determine whether the sequence converges or diverges, and if it converges we are asked to find the limit. We are given the sequence
    a_n = \sqrt[n]{2^{1+3n}}

    So I can see by putting this into my calculator that this is going to converge to 8. My problem is showing this mathematically. Here's where I'm stuck.

    a_n = \sqrt[n]{2^{1+3n}} = \left(2^{1+3n}\right)^{1/n}

    Easy enough, now the tough part....

    \displaystyle \lim_{n \to \infty}\left(2^{1+3n}\right)^{1/n}

    Where do I even begin on this? Do I need to use a natural log or something? Wood that be the thing to do? (terrible pun, but I had to do it.)

    I obviously can't say that my exponent is going to 0 as n goes to infinity, because that's just not true. If it were the answer would just be 1 and that'd be it, but that's obviously not the case.


    If anyone could help point me in the right direction, that'd just be super.

    Thanks!
    Last edited by pirateboy; September 28th 2010 at 02:08 AM. Reason: solved
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by pirateboy View Post
    Here' we're asked to determine whether the sequence converges or diverges, and if it converges we are asked to find the limit. We are given the sequence
    a_n = \sqrt[n]{2^{1+3n}}

    So I can see by putting this into my calculator that this is going to converge to 8. My problem is showing this mathematically. Here's where I'm stuck.

    a_n = \sqrt[n]{2^{1+3n}} = \left(2^{1+3n}\right)^{1/n}

    Easy enough, now the tough part....

    \displaystyle \lim_{n \to \infty}\left(2^{1+3n}\right)^{1/n}

    Where do I even begin on this? Do I need to use a natural log or something? Wood that be the thing to do? (terrible pun, but I had to do it.)

    I obviously can't say that my exponent is going to 0 as n goes to infinity, because that's just not true. If it were the answer would just be 1 and that'd be it, but that's obviously not the case.


    If anyone could help point me in the right direction, that'd just be super.

    Thanks!
    use the properties of exponents This gives

    \displaystyle (2^{1+3n})^{\frac{1}{n}}=2^{\frac{1+3n}{n}}=2^{\fr  ac{1}{n}+3}=2^{\frac{1}{n}}\cdot 2^{3}=8\cdot 2^{\frac{1}{n}}
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  3. #3
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    Hi pirateboy.
    The limit becomes easy if you see
     (2^{1+3n})^{1/n} = e^{ \ln (2^{1/n+3})
    So, by the continuity of e and the properties of ln, the problem becomes to calculate the limit,
    (1/n + 3) \ln (2), When [tex] n \rightarrow \infty [\MATH] you get
    [tex]3\ln(2) = ln(2^3)[tex]. The initial limit yields  e^ln(2^3) =  2^3= 8 .
    I hope this can help you.
    Everk
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  4. #4
    Junior Member pirateboy's Avatar
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    Doh! I can't believe I didn't see that!

    Everk, I think I see what you're talking about.
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  5. #5
    Junior Member pirateboy's Avatar
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    Solved

    So to finish it out...

    \displaystyle \lim_{n \to \infty}\sqrt[n]{2^{1+3n}} = \lim_{n \to \infty}\left(2^{1+3n}\right)^{1/n} =\lim_{n \to \infty} 2^{1/n + 3} = 2^{0+3} = 8

    Thanks, guys.
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