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Math Help - Series questions

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Series questions

    Hey, would anyone mind railing off some very difficult Taylor/Maclaurin series problems, as well as series convergence/divergence problems, and even some limit ones.

    The reason is that I am now able to do all the problems in any book I can find, but I know there are millions of them I probably can't do, and I would love to see them so I can learn, for that is my greatest passion. And I couldn't think of a place that would have more people with knowledge of these kind of problems than MHF.

    And anyways, you get to compete in the MHF sport of embarrassing me


    So any problems of those three subjects would be greatly appreciated, I will solve them and ask for criticism


    Extremely appreciated

    Mathstud
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  2. #2
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    You should learn Taylor's Theorem.
    It is a way to approximate the remainder when you approximate a function with the Taylor polynomial.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You should learn Taylor's Theorem.
    It is a way to approximate the remainder when you approximate a function with the Taylor polynomial.
    Hmm, isn't that just the same as the Lagrange Remainder theorem?
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  4. #4
    Eater of Worlds
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    Try this one to see whether it converges or not.

    \sum_{n=0}^{\infty}\frac{1\cdot{3}\cdot{5}\cdot\cd  ot\cdot(2n-1)}{2^{n}(n+1)!}


    Here's one, similar to the other, you can try Gauss's test on:

    \sum_{n=0}^{\infty}\left[\frac{1\cdot{3}\cdot{5}\cdot\cdot\cdot(2n-1)}{2^{n}n!}\right]^{p}

    Here is also a curious little tidbit:

    What if we take the divergent harmonic series, \sum_{n=1}^{\infty}\frac{1}{n} and remove all terms that have a 9 in them?.

    Is it then convergent?.
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  5. #5
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    Are you really serious you want to learn more about series?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Try this one to see whether it converges or not.

    \sum_{n=0}^{\infty}\frac{1\cdot{3}\cdot{5}\cdot\cd  ot\cdot(2n-1)}{2^{n}(n+1)!}


    Here's one, similar to the other, you can try Gauss's test on:

    \sum_{n=0}^{\infty}\left[\frac{1\cdot{3}\cdot{5}\cdot\cdot\cdot(2n-1)}{2^{n}n!}\right]^{p}
    For the first one

    Well the first one diverges by the n-th term test doesnt it? and the second one I am unsure about it except breaking it up into p>0,p=0,p<0

    And for your last one, intuition says it still diverges, but I am unsure as to how to show it
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  7. #7
    Eater of Worlds
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    The first one converges.

    If you are curious:

    There is a neat corollary from Cauchy which says that if

    {a_{n}} is a positive decreasing sequence, then

    \sum_{n=1}^{\infty}a_{n} converges if

    \lim_{n\rightarrow{\infty}}\frac{a_{2n}}{a_{n}}<\f  rac{1}{2}

    and diverges if

    \lim_{n\rightarrow{\infty}}\frac{a_{2n+1}}{a_{n}}>  \frac{1}{2}

    Look at the relation: \frac{a_{2n}}{a_{n}}=\frac{a_{n+1}}{a_{n}}\cdot\fr  ac{a_{n+2}}{a_{n+1}}\cdot\cdot\cdot\frac{a_{2n}}{a  _{2n-1}}

    \frac{a_{2n+1}}{a_{n}}=\frac{a_{n+1}}{a_{n}}\cdot\  frac{a_{n+2}}{a_{n+1}}\cdot\cdot\cdot\frac{a_{2n+1  }}{a_{2n}}


    Then \frac{a_{2n+1}}{a_{n}}<\frac{a_{2n}}{a_{n}}

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

    =\frac{(2n+3)(2n+5)\cdot\cdot\cdot(4n-1)}{2^{n}(n+2)(n+3)\cdot\cdot\cdot(2n)}

    =\frac{1}{2}\left(\frac{2n+3}{2n+4}\right)\left(\f  rac{2n+5}{2n+6}\right)\cdot\cdot\cdot\left(\frac{4  n-1}{4n}\right)

    <\frac{1}{2}\left(\frac{4n-1}{4n}\right)^{n-1}=\frac{1}{2}\left(1-\frac{1}{4n}\right)^{n-1}

    Since \lim_{n\rightarrow{\infty}}\left(1-\frac{1}{4n}\right)^{n-1}=\frac{1}{e^{\frac{1}{4}}}

    \limsup_{n\rightarrow{\infty}}\frac{a_{2n+1}}{a_{n  }}\leq\limsup_{n\rightarrow{\infty}}\frac{a_{2n}}{  a_{n}}\leq\frac{1}{2e^{\frac{1}{4}}}<\frac{1}{2}

    Therefore, the series converges.


    The second one converges if p>2 and diverges if p\leq{2}

    The third one, the harmonic, surprisingly converges.

    It can be shown that if we delete from the harmonic series all terms that contain X, the resulting series in convergent.
    Last edited by galactus; June 1st 2008 at 09:56 AM. Reason: corollary added
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    The first one converges.

    If you are curious:

    Look at the relation: \frac{a_{2n}}{a_{n}}=\frac{a_{n+1}}{a_{n}}\cdot\fr  ac{a_{n+2}}{a_{n+1}}\cdot\cdot\cdot\frac{a_{2n}}{a  _{2n-1}}

    \frac{a_{2n+1}}{a_{n}}=\frac{a_{n+1}}{a_{n}}\cdot\  frac{a_{n+2}}{a_{n+1}}\cdot\cdot\cdot\frac{a_{2n+1  }}{a_{2n}}


    Then \frac{a_{2n+1}}{a_{n}}<\frac{a_{2n}}{a_{n}}

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

    =\frac{(2n+3)(2n+5)\cdot\cdot\cdot(4n-1)}{2^{n}(n+2)(n+3)\cdot\cdot\cdot(2n)}

    =\frac{1}{2}\left(\frac{2n+3}{2n+4}\right)\left(\f  rac{2n+5}{2n+6}\right)\cdot\cdot\cdot\left(\frac{4  n-1}{4n}\right)

    <\frac{1}{2}\left(\frac{4n-1}{4n}\right)^{n-1}=\frac{1}{2}\left(1-\frac{1}{4n}\right)^{n-1}

    Since \lim_{n\rightarrow{\infty}}\left(1-\frac{1}{4n}\right)^{n-1}=\frac{1}{e^{\frac{1}{4}}}

    \limsup_{n\rightarrow{\infty}}\frac{a_{2n+1}}{a_{n  }}\leq\limsup_{n\rightarrow{\infty}}\frac{a_{2n}}{  a_{n}}\leq\frac{1}{2e^{\frac{1}{4}}}<\frac{1}{2}

    Therefore, the series converges.


    The second one converges if p>2 and diverges if p\leq{2}

    The third one, the harmonic, surprisingly converges.

    It can be shown that if we delete from the harmonic series all terms that contain X, the resulting series in convergent.
    Thanks a lot! That first one, I have done some before, where I used teh ratio test, I should have saw that..very good problems ....any more of those or other types of specified problems would be great!
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    I didnt want to start a new thread, but over all of my series exploration there are only a few that I cant find the divergence or convergence of,

    If anyone could help

    \sum_n\sqrt[n]{a}-1

    \sum_n\sqrt[n]{a}-1-\frac{1}{n} \leftarrow depending on the convergence of the first one this could just be a simple convergent-divergent=divergent one


    \sum_n\frac{1}{(\ln(\ln(n)))^{\ln(n)}}

    and finally

    Find all \alpha and \lambda

    such that the following series is convergent

    \sum_n\bigg[1+\frac{1}{n\ln^{\lambda}(n)}\bigg]^{-n^{\alpha}}

    Those are the pesky ones ...

    I am sure they are very simple, but I just dont see them, could someone enlighten me


    Also any more questions woudl be appreciated

    EDIT: for the first one I think you have to show that \forall{\alpha}>1\text{  }|\sqrt[n]{a}-1|\geq\frac{1}{n}


    But I am not sure
    Last edited by Mathstud28; June 1st 2008 at 01:16 PM.
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  10. #10
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post

    \sum_n\sqrt[n]{a}-1
    \sqrt[n]{a}-1=e^{\frac{\ln(a)}{n}}-1=\frac{\ln(a)}{n}+...
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  11. #11
    Eater of Worlds
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    \sum_{n=2}^{\infty}\frac{1}{(\ln(\ln(n)))^{\ln(n)}  }

    I believe this one is going to be complex.
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    \sum_{n=2}^{\infty}\frac{1}{(\ln(\ln(n)))^{\ln(n)}  }

    I believe this one is going to be complex.
    Haha, we are on a mathwebsite Galactus complex i or complex as in difficult?

    EDIT: I take that back...I dont know waht you maen
    Last edited by Mathstud28; June 1st 2008 at 02:19 PM.
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  13. #13
    Eater of Worlds
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    Complex as the field of complex numbers. I never use complex in reference to difficult because of that reason.
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Complex as the field of complex numbers. I never use complex in reference to difficult because of that reason.
    Really? Thats strange, because the chapter of the book I got it from was called "Positive real sequences"

    Hmm, they probably made a typo...thanks Galactus!
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  15. #15
    Eater of Worlds
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    Maybe I'm wrong, but when I check it I get complex results.
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