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Math Help - rate of convergence

  1. #1
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    rate of convergence

    I want to find the rate of convergence of

    lim as n goes to infinity of (ln(n+1) - ln(n)), which equals
    lim as n goes to infinity of ln((n+1)/n)

    I said that ln[(n+1)/n) must be less than or equal to (n+n)/n^2 and it seems to work. Then, I think the rate if O(1/n).

    However, I'm not sure how exactly to bound the ln function. Can someone show me a bound or why this works?
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    I want to find the rate of convergence of

    lim as n goes to infinity of (ln(n+1) - ln(n)), which equals
    lim as n goes to infinity of ln((n+1)/n)

    I said that ln[(n+1)/n) must be less than or equal to (n+n)/n^2 and it seems to work. Then, I think the rate if O(1/n).

    However, I'm not sure how exactly to bound the ln function. Can someone show me a bound or why this works?
    Note that \log (x+1) \leq x for x>-1 this means \log y \leq y - 1 for y>0.

    Thus, \log \frac{n+1}{n} \leq \frac{n+1}{n} - 1 = \frac{1}{n}.
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