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Math Help - Gauss Test for Convergence

  1. #1
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    Gauss Test for Convergence

    I just got done proving Gauss' test, which is given in the book as:

    If there is an N\ge 1, an s>1, and an M>0 such that
    \frac{a_{n+1}}{a_n}=1 - \frac{A}{n} + \frac{f(n)}{n^s}
    where |f(n)|\le M for all n, then \sum a_n converges if A>1 and diverges if A \le 1.

    This is equivalent to the many other forms I have found on the web. The next question asks to use this test to prove that the series

    \sum_{n=1} ^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)} \right)^k

    converges if k>2 and diverges if k \le 2 using Gauss' test.


    OK, so much for the preamble. Here's my attempt:

    \frac{a_{n+1}}{a_n} = \left ( \frac {2n+1}{2n+2} \right ) ^ k

    And that's it... I don't know how to put this into a form which corresponds in general to the form required for Gauss' test. I saw a similar problem online solved with the use of the fact that \left (\frac{n}{n+1} \right)^k \approx 1 - \frac{k}{n} for large n, but the book has not covered anything like that (it introduced the ~ symbol, but not \approx).
    Last edited by process91; October 5th 2011 at 06:05 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Gauss Test for Convergence

    Quote Originally Posted by process91 View Post
    I just got done proving Gauss' test, which is given in the book as:

    If there is an N\ge 1, an s>1, and an M>0 such that
    \frac{a_{n+1}}{a_n}=1 - \frac{A}{n} + \frac{f(n)}{n^s}
    where |f(n)|\le M for all n, then \sum a_n converges if A>1 and diverges if A \le 1.

    This is equivalent to the many other forms I have found on the web. The next question asks to use this test to prove that the series

    \sum_{n=1} ^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)} \right)^k

    converges if k>2 and diverges if k \le 2 using Gauss' test.


    OK, so much for the preamble. Here's my attempt:

    \frac{a_{n+1}}{a_n} = \left ( \frac {2n+1}{2n+2} \right ) ^ k

    And that's it... I don't know how to put this into a form which corresponds in general to the form required for Gauss' test. I saw a similar problem online solved with the use of the fact that \left (\frac{n}{n+1} \right)^k \approx 1 - \frac{k}{n} for large n, but the book has not covered anything like that (it introduced the ~ symbol, but not \approx).
    Starting from Your result...

    \frac{a_{n+1}}{a_{n}}= (\frac{2n+1}{2n+2})^{k} (1)

    ... we obtain...

    \ln \frac{a_{n+1}}{a_{n}}= k\ \{\ln (1+\frac{1}{2n})-\ln (1+\frac{1}{n})\} = k\ \{-\frac{1}{2n} + \text{o}(\frac{1}{n})\} (2)

    ... so that is...

    \frac{a_{n+1}}{a_{n}}= e^{-k \{\frac{1}{2n}+ \text{o} (\frac{1}{n})\}}= 1 - \frac{k}{2n} + \text{o} (\frac{1}{n}) (3)

    Now applying the Gauss' test we find that the series...

    \sum_{n=1} ^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)} \right)^k (4)

    ... converges for k>2 and diverges for k\le 2...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Gauss Test for Convergence

    Thank you very much for your reply. I'm still a little confused; either the book has not yet covered anything to explain the steps in both (2) and (3), or I missed it (this is Calculus Vol I. by Apostol). It did cover little-o notation in the context of Taylor Series, but that was only useful at the time because we were considering n \rightarrow a. I see that

    \lim_{n \rightarrow \infty} \frac{\ln(1+\frac{1}{2n}) - \ln(1+\frac{1}{n})+\frac{1}{2n}}{\frac{1}{n}}=0, so I understand how the statement is true, but have no idea how to have come up with attempting that on my own. Is this a known result that one will apply? Where could I read more about this?

    In step (3), I am completely at a loss for how to proceed. I don't know what to do with the little-o in the exponent, and I'm not sure how you got your result.

    I think that the answer to this question requires external knowledge about little-o notation that I do not have yet. I don't want to take up so much of your time, if you would be so kind as to point me to another reference I would appreciate it.
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