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Math Help - sequences

  1. #1
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    sequences

    Find the limit of (1+(4/3n+1))^6n.

    I found it as e^6. [(1+(4/3/n+1)/(3))^6]^n and by the equation (1+1/x)^x=e,it is e^6.
    Last edited by kastamonu; February 23rd 2013 at 12:21 PM.
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  2. #2
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    Re: sequences

    (1+4/3n+1)^{6n} = \left(2+\frac{4}{3n}\right)^{6n} > 2^{6n}\to\infty as n\to\infty.

    Secondly, this is not discrete mathematics, but calculus.
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  3. #3
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    Re: sequences

    answer is e^8 but I can't get it.
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  4. #4
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    Re: sequences

    Quote Originally Posted by kastamonu View Post
    Find the limit of (1+(4/3n+1))^6n.
    Sorry, this won't work either.

    (1+(4/3n+1))^{6n} = \left(1+\left(\frac{4}{3n}+1\right)\right)^{6n} = \left(2+\frac{4}{3n}\right)^{6n} > 2^{6n}\to\infty as n\to\infty.

    And there is a question whether 4/3n means (4/3)n or 4/(3n).
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  5. #5
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    Re: sequences

    Quote Originally Posted by kastamonu View Post
    Find the limit of (1+(4/3n+1))^6n.

    I an sure you mean \left( {1 + \frac{4}{{3n + 1}}} \right)^{6n}


    If a\cdot b\ne 0 then \left( {1 + \frac{a}{{bn + c}}} \right)^{dn}  \to e^{\frac{{ad}}{b}} .
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  6. #6
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    Re: sequences

    Yes Plato. Many Thanks. But is there a way to derive this formula?
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  7. #7
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    Re: sequences

    Quote Originally Posted by kastamonu View Post
    Yes Plato. Many Thanks. But is there a way to derive this formula?

    Write it as \left( {1 + \frac{{\frac{a}{b}}}{{n + \frac{c}{b}}}} \right)^{dn} .

    Let \frac{1}{m} = \frac{{\frac{a}{b}}}{{n + \frac{c}{b}}} so n=\tfrac{am}{b}-\tfrac{c}{b}.

    Or \left( {1 + \frac{a}{{bn + c}}} \right)^{dn}  = \left( {\left( {1 + \frac{1}{m}} \right)^m } \right)^{\frac{{ad}}{b}} \left( {1 + \frac{1}{m}} \right)^{ - \frac{c}{b}} .
    Last edited by Plato; February 23rd 2013 at 01:49 PM.
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  8. #8
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    Re: sequences

    Many Thanks.
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