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Math Help - Limit (1)

  1. #16
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    i equally thanked everybody once! PaulRS' solution is nice and my solution differs from his in the last setp only, where i applied Cesaro's mean, as Moo mentioned, to the sequence

    y_n=\frac{1}{x_{n+1}}-\frac{1}{x_n}. that will give you \lim_{n\to\infty} n x_{n+1}=2 and thus \lim_{n\to\infty} (n+1)x_{n+1}=2.
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  2. #17
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    Define y_n=n x_n, then  y_{n+1}=(n+1)\ln(1+\frac {y_n}{n})

    If
    \lim_{n\to\infty}y_n exists, it implies:
    (1)
    \lim_{n\to\infty} \frac {y_n}{n} = 0
    (2) Series z_1=y_1, z_{n+1}=y_{n+1}-y_n converges, since \sum_{i=1}^{n+1} z_i= y_{n+1}.

    z_{n+1}=(n+1) \ln(1+\frac{y_n}{n})-y_n
    =(n+1) \{\ln(1+\frac{y_n}{n})-\frac{y_n}{n}+\frac{y_n^2}{2n^2}\}+\frac{y_n}{2n} \{2-\frac{n+1}{n}y_n\}

    The first term converges since it is O(\frac{1}{n^2}). The second term is \frac{y_n(2-y_n)}{2}\frac{1}{n}+O(\frac{1}{n^2}).

    For the second term to converge, it requires \lim_{n\to\infty}y_n(2-y_n)=0, so \lim_{n\to\infty}y_n=2, noting \lim_{n\to\infty}y_n>0. This also confirms \lim_{n\to\infty}y_n does exist.


    Last edited by mr fantastic; September 19th 2009 at 02:02 AM. Reason: Restored original reply
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