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Math Help - Series converge

  1. #1
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    Thumbs up Series converge

    1-\frac{1}{3}\left(1+\frac{1}{2}\right)+\frac{1}{5}\  left(1+\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{7}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1  }{4}\right)+...

    show that the above series is converges
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  2. #2
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    Looks like somewhat are alternating series to me, find a function to which to fit the pattern of the terms you were given and simply use the alternating series test if you can
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  3. #3
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    Talking

    You probably mean this : (-1)^{2n-1}\frac{\ln n}{2n-1}

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  4. #4
    MHF Contributor chisigma's Avatar
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    The series can be written in compact form as...

    \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{n}}{2n-1} (1)

    ... where...

    H_{n} = \sum_{k=1}^{n} \frac{1}{k} (2)

    It is well known the limit...

    \lim_{n \rightarrow \infty} H_{n} - \ln n = \gamma (3)

    ... where \gamma is the 'Euler's constant'. Consequence of (3) is...

    \lim_{n \rightarrow \infty} \frac{H_{n}}{2n-1} = \lim_{n \rightarrow \infty} \frac{\ln n + \gamma}{2n-1} = 0 (4)

    ... so that the (1) is an alternating decreasing terms series with the general term that in absolute value tends to 0 if n \rightarrow \infty. Therefore (1) is convergent...

    Kind regards

    \chi \sigma
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  5. #5
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    Aw, man I was trying to figure out the function for it, glad I started 5 minutes ago
    That is a new version of the alternating series I have not worked on yet so thank you for the question and thank you to chisigma for the answer.
    Holy, that is a very new series in this form for me I would have never got natural log, how do you know when to apply such things chisigma
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  6. #6
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    Quote Originally Posted by Xingyuan View Post
    1-\frac{1}{3}\left(1+\frac{1}{2}\right)+\frac{1}{5}\  left(1+\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{7}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1  }{4}\right)+...

    show that the above series is converges
    HI

    You can break these to find the general term .

    1/1-1/3+1/5-1/7+... notice that the denominators are odd and the signs alternate so it would be

    \sum^{\infty}_{n=1}[\frac{1}{2n-1}(-1)^{n+1}]

    The other part :

    (1)+(1+1/2)+(1+1/2+1/3)+... The pattern should be obvious here . Every subsequent fraction added has its denominator equal to the no of terms .

    \sum^{\infty}_{n=1}[\sum^{n}_{r=1}(\frac{1}{r})]

    Putting them together :

    \sum^{\infty}_{n=1}[\frac{1}{2n-1}(-1)^{n+1}\cdot \sum^{n}_{r=1}(\frac{1}{r})]

    Notice that as n approaches infinity , \frac{1}{2n-1}(-1)^{n+1} would approach 0 so that this series would eventually converge to 0 .
    Last edited by mathaddict; November 27th 2009 at 08:17 AM.
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  7. #7
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    Quote Originally Posted by mathaddict View Post
    HI

    You can break these to find the general term .

    1/1-1/3+1/5-1/7+... notice that the denominators are odd and the signs alternate so it would be

    \sum^{\infty}_{n=1}[\frac{1}{2n-1}(-1)^{n+1}]

    The other part :

    (1)+(1+1/2)+(1+1/2+1/3)+... The pattern should be obvious here . Every subsequent fraction added has its denominator equal to the no of terms .

    \sum^{\infty}_{r=1}[\sum^{n}_{r=1}(\frac{1}{n})]

    Putting them together :

    \sum^{\infty}_{n=1}[\frac{1}{2n-1}(-1)^{n+1}\cdot \sum^{n}_{r=1}(\frac{1}{r})]

    Notice that as n approaches infinity , \frac{1}{2n-1}(-1)^{n+1} would approach 0 so that this series would eventually converge to 0 .
    Why right here would you say the to the nth term not \infty


    \sum^{\infty}_{r=1}[\sum^{n}_{r=1}(\frac{1}{n})]

    Just curious I am still fairly new to this
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  8. #8
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    Quote Originally Posted by RockHard View Post
    Why right here would you say the to the nth term not \infty


    \sum^{\infty}_{r=1}[\sum^{n}_{r=1}(\frac{1}{n})]

    Just curious I am still fairly new to this
    HI

    My bad . It should be

    \sum^{\infty}_{n=1}[\sum^{n}_{r=1}(\frac{1}{r})]

    \sum^{\infty}_{n=1}=\sum^{1}_{r=1}(\frac{1}{r})+\s  um^{2}(\frac{1}{r})+\sum^{3}_{r=1}(\frac{1}{r})+..  .
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