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Math Help - A limit of a sequence \ series

  1. #1
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    A limit of a sequence \ series

    Prove that the sequence (why do they call it a sequence?):
    \sum_{k=n}^{3n}\frac{1}{k} has a limit 'a', which is in the range [ \frac{2}{3},\frac{3}{2}]

    I don't even know how to look at it : as a series, or as a sequence. I know it should be really easy, and should use some plain lemmas or convergence tests, I just don't know which one.

    Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    From the well known relation...

    \lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{1}{k} - \ln n = \gamma (1)

    ... where \gamma is the so called 'Euler's constant' we derive...

    \lim_{n \rightarrow \infty} \sum_{k=1}^{3n}\frac{1}{k} - \ln 3n <br />
=\lim_{n \rightarrow \infty} \sum_{k=1}^{n-1}\frac{1}{k} - \ln (n-1) = \gamma (2)

    ... and because is...

    \sum_{k=n}^{3n}\frac{1}{k} = \sum_{k=1}^{3n}\frac{1}{k} - \sum_{k=1}^{n-1}\frac{1}{k} (3)

    ... we have...

    \lim_{n \rightarrow \infty} \sum_{k=n}^{3n}\frac{1}{k} = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{3n}\frac{1}{k} - \sum_{k=1}^{n-1}\frac{1}{k} \}=

     \lim_{n \rightarrow \infty} \{ \ln 3n - \ln (n-1)\} = \lim_{n \rightarrow \infty} \ln \frac{3n}{n-1} = \ln 3 = 1.09861228866 ... (4)

    Kind regards

    \chi \sigma
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  3. #3
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    Wow, that's amazing! We haven't learned that yet, but it looks very interesting!

    But unfortunately I can't use this, I need to use more plain techniques for proving this...

    Thank you!
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by adam63 View Post
    Prove that the sequence (why do they call it a sequence?):
    \sum_{k=n}^{3n}\frac{1}{k} has a limit 'a', which is in the range [ \frac{2}{3},\frac{3}{2}]

    I don't even know how to look at it : as a series, or as a sequence. I know it should be really easy, and should use some plain lemmas or convergence tests, I just don't know which one.

    Thanks!
    It's a sequence:


    S_n=\sum_{k=n}^{3n}\frac{1}{k}

    How you proceed depends on what you know and/or what you were doing in class at the time this was set. I would trap S_n between a pair of integrals and using the squeeze theorem show that:

    \lim_{n \to \infty}S_n=\ln(3)

    CB
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  5. #5
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    Why is it a sequence? It looks as if a_k=\frac{1}{k} is the sequence, and the sum of it is the series.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by adam63 View Post
    Why is it a sequence? It looks as if a_k=\frac{1}{k} is the sequence, and the sum of it is the series.
    The sums form a sequence.

    For an infinite series the sum if it exists is the limit of the sequence of partial sums.

    CB
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  7. #7
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    the lower bound is obvious :


     S = \frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n-1} + \frac{1}{3n}


    since

     \frac{1}{n} > \frac{1}{n+1} > ... > \frac{1}{3n}

    the series is greater than

     (2n+1)  \frac{1}{3n}

    if n tends to infinity ,

     S > \frac{2}{3}



    For the upper bound ,


      S = \frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n-1} + \frac{1}{3n}

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

     = \frac{4n}{n(3n)} + \frac{4n}{(n+1)(3n-1)} + ... + \frac{4n}{(2n-1)(2n+1)} + \frac{1}{2n}

    Consider

     (n+x)(3n-x) = -x^2 + 2nx + 3n^2

     = -(x-n)^2 + 4n^2

    we can see that for  0 \leq x \leq n-1

     (n+x)(3n-x) is min. when  x=0

     \frac{1}{n(3n)} is max.

    therefore ,  S < (4n) \frac{n}{n(3n) } + \frac{1}{2n} = \frac{4}{3} + \frac{1}{2n}

     \frac{4}{3} could be the upper bound which is better than 1.5
    Last edited by simplependulum; December 4th 2009 at 04:14 AM.
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  8. #8
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    Quote Originally Posted by chisigma View Post
    From the well known relation...

    \lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{1}{k} - \ln n = \gamma (1)

    ... where \gamma is the so called 'Euler's constant' we derive...

    \lim_{n \rightarrow \infty} \sum_{k=1}^{3n}\frac{1}{k} - \ln 3n <br />
=\lim_{n \rightarrow \infty} \sum_{k=1}^{n-1}\frac{1}{k} - \ln (n-1) = \gamma (2)

    ... and because is...

    \sum_{k=n}^{3n}\frac{1}{k} = \sum_{k=1}^{3n}\frac{1}{k} - \sum_{k=1}^{n-1}\frac{1}{k} (3)

    ... we have...

    \lim_{n \rightarrow \infty} \sum_{k=n}^{3n}\frac{1}{k} = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{3n}\frac{1}{k} - \sum_{k=1}^{n-1}\frac{1}{k} \}=

     \lim_{n \rightarrow \infty} \{ \ln 3n - \ln (n-1)\} = \lim_{n \rightarrow \infty} \ln \frac{3n}{n-1} = \ln 3 = 1.09861228866 ... (4)

    Kind regards

    \chi \sigma


    hi

    i guess you love Euler's constant very much because i can always see how you make use of it


    <br />
\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{1}{k} - \ln n = \gamma<br />

    It is another special constant besides  \pi and  e . Nice indeed !
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  9. #9
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    Quote Originally Posted by simplependulum View Post

     \frac{1}{n} > \frac{1}{n+1} > ... > \frac{1}{3n}

    the series is greater than

     (2n+1)  \frac{1}{3n}

    Why is that?

    And if a series is bounded, then it's still not enough for saying that it has a limit.
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  10. #10
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    Quote Originally Posted by adam63 View Post
    Why is that?

    And if a series is bounded, then it's still not enough for saying that it has a limit.

    I could just explain your first question , other is left to some experts


     S = \frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n-1} + \frac{1}{3n}

    we know that there is  3n - (n) + 1 = 2n+1 terms here .

    and since  \frac{1}{n} > \frac{1}{n+1} > \frac{1}{n+2} >... > \frac{1}{3n}

    therefore

     S > \frac{1}{3n} + \frac{1}{3n} + ... + \frac{1}{3n}

    There are totally 2n+1 terms so

     S > (2n+1)\frac{1}{3n}
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  11. #11
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    for each n\le k\le3n we have \frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n}, hence \sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}}, and \frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n}, finally as n\longrightarrow\infty you'll get the lower bound \frac23 and upper bound 2, which is a better result that you asked!
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  12. #12
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    Quote Originally Posted by Krizalid View Post
    for each n\le k\le3n we have \frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n}, hence \sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}}, and \frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n}, finally as n\longrightarrow\infty you'll get the lower bound \frac23 and upper bound 2, which is a better result that you asked!

    Hi ,

    adam63 asked how to prove there exists a limit which is bounded . I think it is obvious but i do not know how to complete this proof ...
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by Krizalid View Post
    for each n\le k\le3n we have \frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n}, hence \sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}}, and \frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n}, finally as n\longrightarrow\infty you'll get the lower bound \frac23 and upper bound 2, which is a better result that you asked!
    The upper bound asker for is 3/2 which is tighter than 2.

    CB
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  14. #14
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    ah yes, i got my mind in clouds.
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