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Thread: limit of (2^n)/(n(n+1)(n+2))

  1. #1
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    limit of (2^n)/(n(n+1)(n+2))

    How would I know that the limit of (2^n)/(n(n+1)(n+2)) as n tends to infinity is 0?
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    Re: limit of (2^n)/(n(n+1)(n+2))

    It isn't. It diverges to infinity.
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    Re: limit of (2^n)/(n(n+1)(n+2))

    WolframAlpha says it is 0.

    (2^n)/(n(n+1)(n+2)) - Wolfram|Alpha
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    Re: limit of (2^n)/(n(n+1)(n+2))

    Quote Originally Posted by Stuck Man View Post
    WolframAlpha says it is 0.

    (2^n)/(n(n+1)(n+2)) - Wolfram|Alpha
    No it does not
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    Re: limit of (2^n)/(n(n+1)(n+2))

    Hello, Stuck Man!

    How would I know that the limit of (2^n)/(n(n+1)(n+2)) as n tends to infinity is 0?

    I don't understand how Wolfram came up with that . . . but it's wrong!

    An exponential function will outgrow a cubic function.



    Darn! . . . LaTex isn't working!!


    . . . . . . . . . . . . . . .2048
    At n = 11, we have: ------
    . . . . . . . . . . . . . . .1716

    . . . . . . . . . . . . . . .1,048,576
    At n = 20, we have: -----------
    . . . . . . . . . . . . . . . . 9,240
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    Re: limit of (2^n)/(n(n+1)(n+2))

    Wolframalpha says the limit as n tends to NEGATIVE infinity is zero (which is true). The numerator becomes arbitrarily close to zero while the absolute value of the denominator grows arbitrarily large.
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  7. #7
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    Re: limit of (2^n)/(n(n+1)(n+2))

    Quote Originally Posted by Stuck Man View Post
    WolframAlpha says it is 0.

    (2^n)/(n(n+1)(n+2)) - Wolfram|Alpha
    The first graph given is misleading, the second graph (which is zoomed out) gives a better picture...
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    Re: limit of (2^n)/(n(n+1)(n+2))

    I misread WolframAlpha. It gives the limit as n tends to negative infinity. If its true that in general an exponential function outgrows a cubic function or one with a higher degree then I will know what the limit is in future. That's assuming the base is greater than 1.

    The graphs omit the growth pattern that starts getting noticeable at about x=8.
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    Re: limit of (2^n)/(n(n+1)(n+2))

    Quote Originally Posted by Stuck Man View Post
    I misread WolframAlpha. It gives the limit as n tends to negative infinity. If its true that in general an exponential function outgrows a cubic function or one with a higher degree then I will know what the limit is in future. That's assuming the base is greater than 1.

    The graphs omit the growth pattern that starts getting noticeable at about x=8.
    Let p be any polynomial such that the Limit as n approaches infinity of p(n) is infinity. Then the Limit as n approaches infinity of a^n/p(n) is infinity for any a>1.
    Last edited by SlipEternal; Jan 30th 2014 at 12:53 PM.
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