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Math Help - Limit of a Sequence

  1. #1
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    Limit of a Sequence

    apparently every computer program gives me an error finding the limit of this:

    [\frac{4n+1}{n^2}+(\frac{1}{5})^n * \frac{n}{2n+1} + (1+\frac{10}{n})^{3n}

    my result is e^{1000}. looks way too strange for me
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  2. #2
    MHF Contributor chisigma's Avatar
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    The sequence is the sum of three sequences...

    a) \frac {1+4n}{n^2} for which is \lim_{n\rightarrow \infty} =0

    b) (\frac {1}{5})^{n}\cdot \frac{n}{1+2n} for which is \lim_{n\rightarrow \infty} =0

    c) (1+\frac {10}{n})^{3n} for which is \lim_{n\rightarrow \infty} (1+\frac {30}{3n})^{3n}= e^{30}

    ... so that the whole limit is e^{30} ...

    Kind regards

    \chi \sigma
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  3. #3
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    hm that c part is a bit unclear for me

    i thought i could write \lim(1+\frac{10}{n})^{3n} = \lim (1+\frac{10}{n})^n*\lim (1+\frac{10}{n})^n*\lim (1+\frac{10}{n})^n


    oh damn.... i did a really embarassing mistake basic algebra here i come again... e^{10}*e^{10}*e^{10} = e^{30} and not to the power of 1000...

    but thanks, needed to clarify if my thinking was right
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  4. #4
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    One way to solve e sequences is to use this formula:

    (An)^(Bn) = (Bn)(An-1) = L -> Limit is e^L

    (1 + 10/n)^(3n) = 3n(1 + 10/n -1) = 30 -> Limit is e^30.
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