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Math Help - Sequences, convergence

  1. #1
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    Sequences, convergence

    I need to show if the following sequence converges and its limit.

    3n^2+4^n+2^n/n^23^n+4n^7

    The dominant term is 4^n.

    Dividing both numerator and denominator by 4^n, we get

    3n^2/4^n + 1 + (2/4)^n / n^2(3/4)^n + 4n^7/4^n

    Since n^2/4^n, (2/4)^n, n^2(3/4)^n and n^7/4^n are basic null sequences, we find, by the combination rules, that

    lim = 0+1+0/0+0 = ?

    not sure what to do. Can anyone help?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Sequences, convergence

    Is the limit:
    \lim_{x\to \infty} \frac{3n^2+4^n+2^n}{(n^{23})^n+4n^7}
    ?
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  3. #3
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    Re: Sequences, convergence

    Sorry the it should read n^2*3^n not (n^23)^n.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Sequences, convergence

    modified.
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  5. #5
    Ted
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    Re: Sequences, convergence

    \displaystyle \lim_{n\to\infty} \frac{3n^2+4^n+2^n}{n^2 \, 3^n +4n^7}

    Start by dividing top & bottom by 4^n
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  6. #6
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    Re: Sequences, convergence

    I did this and got 3n^2/4^n + 1 + (2/4)^n. I am not sure what to do next.
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  7. #7
    Ted
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    Re: Sequences, convergence

    You should get : \displaystyle \lim_{n\to\infty} \frac{ \frac{3n^2}{4^n} + 1 + \left( \frac{1}{2} \right)^n } { n^2 \left( \frac{3}{4} \right)^n + \frac{4n^7}{4^n} }



    You need two things here :

    (1) L'Hospital's Rule.

    (2) \lim_{n\to\infty} r^n = 0 if |r| < 1.
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