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Thread: 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?):
    $\displaystyle \sum_{k=n}^{3n}\frac{1}{k}$ has a limit 'a', which is in the range [$\displaystyle \frac{2}{3},\frac{3}{2}$]

    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...

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

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

    $\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{3n}\frac{1}{k} - \ln 3n
    =\lim_{n \rightarrow \infty} \sum_{k=1}^{n-1}\frac{1}{k} - \ln (n-1) = \gamma$ (2)

    ... and because is...

    $\displaystyle \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...

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

    $\displaystyle \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

    $\displaystyle \chi$ $\displaystyle \sigma$
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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?):
    $\displaystyle \sum_{k=n}^{3n}\frac{1}{k}$ has a limit 'a', which is in the range [$\displaystyle \frac{2}{3},\frac{3}{2}$]

    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:


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

    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 $\displaystyle S_n$ between a pair of integrals and using the squeeze theorem show that:

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

    CB
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  5. #5
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    Why is it a sequence? It looks as if $\displaystyle a_k=\frac{1}{k}$ 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 $\displaystyle a_k=\frac{1}{k}$ 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 :


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


    since

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

    the series is greater than

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

    if n tends to infinity ,

    $\displaystyle S > \frac{2}{3} $



    For the upper bound ,


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

    $\displaystyle = \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} $

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

    Consider

    $\displaystyle (n+x)(3n-x) = -x^2 + 2nx + 3n^2 $

    $\displaystyle = -(x-n)^2 + 4n^2 $

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

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

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

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

    $\displaystyle \frac{4}{3} $ could be the upper bound which is better than 1.5
    Last edited by simplependulum; Dec 4th 2009 at 03:14 AM.
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  8. #8
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    Quote Originally Posted by chisigma View Post
    From the well known relation...

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

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

    $\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{3n}\frac{1}{k} - \ln 3n
    =\lim_{n \rightarrow \infty} \sum_{k=1}^{n-1}\frac{1}{k} - \ln (n-1) = \gamma$ (2)

    ... and because is...

    $\displaystyle \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...

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

    $\displaystyle \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

    $\displaystyle \chi$ $\displaystyle \sigma$


    hi

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


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

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

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

    the series is greater than

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

    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


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

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

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

    therefore

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

    There are totally $\displaystyle 2n+1$ terms so

    $\displaystyle S > (2n+1)\frac{1}{3n} $
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  11. #11
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    for each $\displaystyle n\le k\le3n$ we have $\displaystyle \frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n},$ hence $\displaystyle \sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}},$ and $\displaystyle \frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n},$ finally as $\displaystyle n\longrightarrow\infty$ you'll get the lower bound $\displaystyle \frac23$ and upper bound $\displaystyle 2,$ 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 $\displaystyle n\le k\le3n$ we have $\displaystyle \frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n},$ hence $\displaystyle \sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}},$ and $\displaystyle \frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n},$ finally as $\displaystyle n\longrightarrow\infty$ you'll get the lower bound $\displaystyle \frac23$ and upper bound $\displaystyle 2,$ 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 $\displaystyle n\le k\le3n$ we have $\displaystyle \frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n},$ hence $\displaystyle \sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}},$ and $\displaystyle \frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n},$ finally as $\displaystyle n\longrightarrow\infty$ you'll get the lower bound $\displaystyle \frac23$ and upper bound $\displaystyle 2,$ 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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