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Math Help - sequence and convergence

  1. #1
    Super Member dhiab's Avatar
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    sequence and convergence

    Knowing that :
    prove that sequence definite by :

    converge to a limit real of which they ask to calculate.
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  2. #2
    MHF Contributor chisigma's Avatar
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    From the 'duoble inequality'...

    \int_{k}^{k+1} \frac{dx}{x} \le \frac{1}{k} \le \int_{k-1}^{k} \frac{dx}{x} (1)

    ... we derive...

    \sum_{k=n+1}^{2n} \int_{k}^{k+1} \frac{dx}{x} \le \sum_{k=n+1}^{2n} \frac{1}{k} \le \sum_{k=n+1}^{2n} \int_{k-1}^{k} \frac{dx}{x} (2)

    ... so that is...

    \int_{n+1}^{2n} \frac{dx}{x} \le u_{n} \le \int_{n}^{2n-1} \frac{dx}{x} (3)

    ... so that is...

    \ln \frac{2n}{n+1} \le u_{n} \le \ln \frac{2n-1}{n} (4)

    From (4) is easy to conclude that is...

    \lim_{n \rightarrow \infty} u_{n} = \ln 2 (5)

    Kind regards

     \chi \sigma
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