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Math Help - Help with series

  1. #1
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    Help with series

    Determine if the following are absolutely convergent, conditionally convergent or divergent.

    (1) \sum_{n=0}^{\infty }{\frac { 1!+2!+...+n!}{ \left(2n \right)!}}

    I am really quite confused as to how to work this out. I would be grateful if someone could point me in the right direction by telling me where I've gone wrong.

    I've used the ratio test and I have a limit that still has factorials in it. But I don't know any other test that would be able to help me get rid of the factorials.
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  2. #2
    Member Nacho's Avatar
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    first whit testo comparation

    <br />
\sum\limits_{n = 1}^\infty  {\frac{{1! + 2! + 3! + ... + n!}}<br />
{{\left( {2n} \right)!}}}  \leqslant \sum\limits_{n = 1}^\infty  {\frac{{n! + n! + n! + ... + n!}}<br />
{{\left( {2n} \right)!}}}  = \sum\limits_{n = 1}^\infty  {\frac{{n \cdot n!}}<br />
{{\left( {2n} \right)!}}} <br />

    If <br />
a_n  = \frac{{n \cdot n!}}<br />
{{\left( {2n} \right)!}}<br />
now I use the quotien test

    <br />
\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}<br />
{{a_n }}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right) \cdot \left( {n + 1} \right)!}}<br />
{{\left( {2(n + 1)} \right)!}} \cdot \frac{{\left( {2n} \right)!}}<br />
{{n \cdot n!}} = <br />
<br />
\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)\left( {n + 1} \right) \cdot n!}}<br />
{{\left( {2n + 2} \right)\left( {2n + 1} \right)\left( {2n} \right)!}} \cdot \frac{{\left( {2n} \right)!}}<br />
{{n \cdot n!}}<br />

    <br />
 = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)\left( {n + 1} \right)}}<br />
{{\left( {2n + 1} \right)\left( {2n + 1} \right)}} \cdot \frac{1}<br />
{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {1 + \frac{1}<br />
{n}} \right)\left( {1 + \frac{1}<br />
{n}} \right) \cdot \frac{1}<br />
{n}}}<br />
{{\left( {2 + \frac{2}<br />
{n}} \right)\left( {2 + \frac{1}<br />
{n}} \right)}}<br />
<br />
 = \frac{{\left( {1 + 0} \right)^2  \cdot 0}}<br />
{{\left( {2 + 0} \right)^2 }} = 0 < 1<br />

    So, how <br />
\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}<br />
{{a_n }}} \right| < 1 \Rightarrow <br />
the serie converge

    I forget say you that if this serie converge, then (1) converge too. And how converge whit and whitout absolute value, the serie is absolutely convergent
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Hweengee View Post
    Determine if the following are absolutely convergent, conditionally convergent or divergent.

    (1) \sum_{n=0}^{\infty }{\frac { 1!+2!+...+n!}{ \left(2n \right)!}}

    I am really quite confused as to how to work this out. I would be grateful if someone could point me in the right direction by telling me where I've gone wrong.

    I've used the ratio test and I have a limit that still has factorials in it. But I don't know any other test that would be able to help me get rid of the factorials.
    There isn't even a need to make a comparison

    \lim_{n\to\infty}\left(\frac{1!+2!+\cdots{+n!}}{(2  n)!}\right)^{\frac{1}{n}}\sim\lim_{n\to\infty}\lef  t(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}=0

    Therefore the serise converges
    Last edited by Mathstud28; November 2nd 2008 at 11:15 AM.
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  4. #4
    Moo
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    What's fun with you Mathstud is that everything is obvious, clear or simple and everybody knows the limits you know


    I don't doubt your limit \lim_{n\to\infty}\left(\frac{1}{(2n)!}\right)^{\fr  ac{1}{n}}=0 is correct, however, I've been taught 0^0 is not defined (by convention, it's 1), so it is not that obvious.

    A brief reminder of Stirling's approximation formula would be nice.

    Tsssk, try to imagine that some people here are not as skilled as you are, please.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    What's fun with you Mathstud is that everything is obvious, clear or simple and everybody knows the limits you know


    I don't doubt your limit \lim_{n\to\infty}\left(\frac{1}{(2n)!}\right)^{\fr  ac{1}{n}}=0 is correct, however, I've been taught 0^0 is not defined (by convention, it's 1), so it is not that obvious.

    A brief reminder of Stirling's approximation formula would be nice.

    Tsssk, try to imagine that some people here are not as skilled as you are, please.
    I am really sorry. I will try to be more dilligent. I can supply the proof if anyone would like it is really interesting but we have that

    n!\sim\sqrt{2\pi{n}}n^ne^{-n}

    What this means is that as n approaches infinity the functions become equivalent.

    From there we can direct substitute to get

    (an)!\sim\sqrt{2a\pi{n}}(an)^{an}e^{-an}=\sqrt{2a\pi{n}}a^{an}\left(n^n\right)^a(e^{-n})^a

    So we can see that

    (2n)!\sim\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2

    So

    \lim_{n\to\infty}\left(\frac{n!}{(2n)!}\right)^{\f  rac{1}{n}}\sim\lim_{n\to\infty}\left(\frac{\sqrt{2  \pi{n}}n^ne^{-n}}{\sqrt{4\pi{n}}(n^n)^2(e^{-n})^2}\right)^{\frac{1}{n}}

    =\lim_{n\to\infty}\left(\frac{1}{\sqrt{2}n^ne^{-n}}\right)^{\frac{1}{n}}

    =\lim_{n\to\infty}\frac{e}{2^{\frac{1}{2n}}n}\to{0  }

    And since

    0\leq\frac{1}{(2n)!}\leq\frac{n!}{(2n)!}

    It can be seen pretty easily by the squeeze theorem that the limit of the middle term is zero as well


    I hope that suffices.
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  6. #6
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    This is an interesting factoring exercise.
    \left( {2n} \right)! = 2^n \left( {n!} \right)(2n - 1)(2n - 3) \cdots (3)(1)

    \frac{{\sum\limits_{k = 1}^n {k!} }}{{\left( {2n} \right)!}} \leqslant \frac{n}{{2^n }}
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  7. #7
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    haha the thing is ~ denotes asymptotic equivalence? something which i have not even learnt. thanks anyway.
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    I am really sorry. I will try to be more dilligent. I can supply the proof if anyone would like it is really interesting but we have that

    n!\sim\sqrt{2\pi{n}}n^ne^{-n}

    What this means is that as n approaches infinity the functions become equivalent.

    From there we can direct substitute to get

    (an)!\sim\sqrt{2a\pi{n}}(an)^{an}e^{-an}=\sqrt{2a\pi{n}}a^{an}\left(n^n\right)^a(e^{-n})^a

    So we can see that

    (2n)!\sim\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2

    So

    \lim_{n\to\infty}\left(\frac{n!}{(2n)!}\right)^{\f  rac{1}{n}}\sim\lim_{n\to\infty}\left(\frac{\sqrt{2  \pi{n}}n^ne^{-n}}{\sqrt{4\pi{n}}(n^n)^2(e^{-n})^2}\right)^{\frac{1}{n}}

    =\lim_{n\to\infty}\left(\frac{1}{\sqrt{2}n^ne^{-n}}\right)^{\frac{1}{n}}

    =\lim_{n\to\infty}\frac{e}{2^{\frac{1}{2n}}n}\to{0  }

    And since

    0\leq\frac{1}{(2n)!}\leq\frac{n!}{(2n)!}

    It can be seen pretty easily by the squeeze theorem that the limit of the middle term is zero as well


    I hope that suffices.
    Small problem:

    \left(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}

    is not the n-th root of the n-th term of the series, so there is still a gap to fill in your argument.

    You need to show that considering the above is equivalent to considering:

    \left[ \frac { 1!+2!+...+n!}{ \left(2n \right)!} \right] ^{1/n}

    CB
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Small problem:

    \left(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}

    is not the n-th root of the n-th term of the series, so there is still a gap to fill in your argument.

    You need to show that considering the above is equivalent to considering:

    \left[ \frac { 1!+2!+...+n!}{ \left(2n \right)!} \right] ^{1/n}

    CB
    Yes but they are asymptotically equivalent. I will come back tonight and show explicitly, but if you are looking for proof of that taking the logaritim of the limit makes this more obvious

    Or just see

    \lim_{n\to\infty}\left(\frac{1!+2!+\cdots+n!}{(2n)  !}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\left(n!\  cdot\frac{\frac{1}{n!}+\frac{2!}{n!}+\cdots+1}{(2n  )!}\right)^{\frac{1}{n}}

    =\lim_{n\to\infty}\left(\frac{1}{n!}+\frac{2!}{n!}  +\cdots+1\right)^{\frac{1}{n}}\cdot\lim_{n\to\inft  y}\left(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}=1\cd  ot{0}=0
    Last edited by Mathstud28; November 3rd 2008 at 03:20 AM.
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    Yes but they are asymptotically equivalent. I will come back tonight and show explicitly, but if you are looking for proof of that taking the logaritim of the limit makes this more obvious

    Or just see

    \lim_{n\to\infty}\left(\frac{1!+2!+\cdots+n!}{(2n)  !}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\left(n!\  cdot\frac{\frac{1}{n!}+\frac{2!}{n!}+\cdots+1}{(2n  )!}\right)^{\frac{1}{n}}

    =\lim_{n\to\infty}\left(\frac{1}{n!}+\frac{2!}{n!}  +\cdots+1\right)^{\frac{1}{n}}\cdot\lim_{n\to\inft  y}\left(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}=1\cd  ot{0}=0
    I didn't say it could not be done, just that you have to do it. but it leaves this as a very clumsy method when you put in the detail you have tried to wave your hands over.

    After all hand waving is a pointless activity with this series, since it is obviously convergent we could just have made that claim if we were going to handwave about it.

    CB
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    I didn't say it could not be done, just that you have to do it. but it leaves this as a very clumsy method when you put in the detail you have tried to wave your hands over.

    After all hand waving is a pointless activity with this series, since it is obviously convergent we could just have made that claim if we were going to handwave about it.

    CB
    Sorry if I did something improper, In all honesty it isn't that laborious in my opinion. As for the hand-waving I totally understand where your coming from but for replying on a math forum it gives a solution but without an integral part, left for the poster. This helps them learn something. And I was just giving an alternate solution. I don't know, that's just my opinion. Thank you for your input.
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