Page 2 of 2 FirstFirst 12
Results 16 to 17 of 17

Thread: Limit (1)

  1. #16
    MHF Contributor

    May 2008
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Aug 2009
    Define y_n=n x_n, then  y_{n+1}=(n+1)\ln(1+\frac {y_n}{n})

    \lim_{n\to\infty}y_n exists, it implies:
    \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; Sep 19th 2009 at 02:02 AM. Reason: Restored original reply
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Replies: 12
    Last Post: Aug 26th 2010, 11:59 AM
  2. Replies: 1
    Last Post: Aug 8th 2010, 12:29 PM
  3. Replies: 1
    Last Post: Feb 5th 2010, 04:33 AM
  4. Replies: 16
    Last Post: Nov 15th 2009, 05:18 PM
  5. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 3rd 2009, 06:05 PM

Search Tags

/mathhelpforum @mathhelpforum