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Math Help - limit, sequence, exists

  1. #1
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    limit, sequence, exists

    Define a_n = 1+\frac{1}{2}+\frac{1}{3}+ \cdots + \frac{1}{n} - \int^n_1 \frac{1}{t}dt. Prove that a=\lim_{n \rightarrow \infty} a_n exists and 0<a<1.
    Attempt
    My issue is the integral in the sequence. I don't know how to take the limit of the sequence with that in there. I think there is a way to break up the integral to take the limit of the sequence easier. I need help with this question. Thanks in advance.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let's indicate with \gamma the limit...

    \gamma = \lim_{n\rightarrow \infty} 1 + \frac {1}{2} + \frac{1}{3} + \dots + \frac{1}{n} - \int_{1}^{n}\frac{dt}{t} = \lim_{n\rightarrow \infty} \int_{1}^{n} (\frac {1}{\lfloor t \rfloor} - \frac{1}{t})\cdot dt = \lim_{n\rightarrow \infty} \sum_{k=1}^{n} \int_{k}^{k+1} (\frac {1}{\lfloor t \rfloor} - \frac{1}{t})\cdot dt

    ... where \lfloor * \rfloor is the so called 'floor function'. On the basis of geometrical considerations [a diagram would be useful to better undestand...] you can demonstrate that is...

    \int_{k}^{k+1} (\frac {1}{\lfloor t \rfloor} - \frac{1}{t})\cdot dt < \frac {1}{2\cdot k^{2}}

    ... so that...

    \gamma < \lim_{n\rightarrow \infty} \frac{1}{2}\cdot \sum_{k=1}^{n} \frac{1}{k^{2}}= \frac{\pi^{2}}{12} = .822467033424 \dots

    Kind regards

    \chi \sigma
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