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Math Help - Limit of sequence

  1. #1
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    Limit of sequence

    \lim_{n\to\infty}n-n^{2}log\frac{n+1}{n}
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  2. #2
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    Found it

    Sorry for occupying the space, but the idea came to me when solving other limit.
    Maybe it will help someone else.

    The initial limit can be written as:
    \lim_{n\to\infty}\frac{\frac{1}{n}-log(1+\frac{1}{n})}{\frac{1}{n^{2}}}
    We can substitute 1/n with x, x real, and we get:
    \lim_{x\to0}\frac{x-log(1+x)}{x^{2}}
    After applying 2 times l'Hopital's rule we get the final answer 1/2.

    I believe there is another solution without l'Hopital and functions, only with sequences .
    Last edited by m3th0dman; December 19th 2009 at 03:37 AM.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Note that for arbitrarily large n it is true that \ln\left(\frac{n+1}{n}\right)=\ln\left(1+\frac{1}{  n}\right)\underset{n\to\infty}{\sim}\frac{1}{n}-\frac{1}{2n^2} so then \lim_{n\to\infty}\left\{n-n^2\ln\left(\frac{n+1}{n}\right)\right\}=\lim_{n\t  o\infty}\left\{n-n^2\left(\frac{1}{n}-\frac{1}{2n^2}\right)\right\}=\frac{1}{2}.
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