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Math Help - Tell if the sequence is bounded or not

  1. #1
    MHF Contributor alexmahone's Avatar
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    Tell if the sequence is bounded or not

    Tell if the sequence is bounded or not (indicate a reason):

    a_n=1+\frac{1}{4}+\frac{1}{7}+\ldots+\frac{1}{3n+1  }
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Tell if the sequence is bounded or not

    My solution:

    a_n>\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\cdots+ \frac{1}{3n+3}

    =\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots  +\frac{1}{n+1}\right)

    =\frac{H_{n+1}}{3}

    Since the harmonic series is unbounded, \{a_n\} is unbounded.
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  3. #3
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    Re: Tell if the sequence is bounded or not

    I agree. The series \sum_{k=0}^\infty} \frac{1}{3k + 3} = \lim_{n \to \infty} \sum_{k = 0}^n \frac{1}{3k + 3} = \lim_{n \to \infty} a_n

    certainly diverges and hence the sequence is unbounded.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Tell if the sequence is bounded or not

    As explained in...

    http://www.mathhelpforum.com/math-he...-i-188482.html

    ... is...

    a_{n}=1+\sum_{k=1}^{n} \frac{1}{3n+1}= 1+ \frac{1}{3}\ \sum_{k=1}^{n} \frac{1}{n+\frac{1}{3}}=

    = 1-\frac{1}{3}\ \phi(\frac{1}{3}) + \frac{1}{3}\ \phi(n+\frac{1}{3}) \sim 1-\frac{1}{3}\ \phi(\frac{1}{3}) + \frac{1}{3}\ \ln (n+\frac{1}{3}) (1)

    ... where...

    \phi(x)= \frac {d}{dx} \ln x! (2)

    Kind regards

    \chi \sigma
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