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Math Help - investigate the convergence at the radius of convergence

  1. #1
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    investigate the convergence at the radius of convergence

    \displaystyle \sum_{n = 0}^{\infty} \displaystyle \frac{1.5.9...(4n+1)}{3.5.7...(2n+3)}x^n


    so i have calculated R=+or-1/6, and when R=1/6, the series reduced to  \displaystyle \frac{1}{3.7.11.15...(4n+3)} (\displaystyle \frac{1}{6})^n, now how do we test if this series convergent or not?


    The second one

    \displaystyle \sum_{n = 4}^{\infty} \displaystyle \frac{(n-1)!(n-4)!}{[7.10.13...(3n-2)]^2}x^{2n}

    for this one, I have got R=+or- 9/7, and the series reduced to \displaystyle \frac{(n-1)!(n-4)!}{[7.10.13...(3n-2)]^2}(9/7)^{2n}, what is the next step?
    Last edited by wopashui; March 25th 2011 at 07:27 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    how did you get R=1/6?

    I thought it was 1/2 as well, but I did this really quick.
    I wanted to see the poster's work.
    Last edited by matheagle; March 25th 2011 at 09:57 PM.
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  3. #3
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    1. Your series is \displaystyle \sum_{n = 0}^{\infty}\left[\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n\right].

    Using the ratio test

    \displaystyle \lim_{n \to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)[4(n+1)+1]}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)[2(n+1)+3]}x^{n+1}}{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n}\right|

    \displaystyle = \lim_{n\to \infty}\left|\left[\frac{4(n+1)+1}{2(n+1)+3}\right]x\right|

    \displaystyle = \lim_{n \to \infty}\left|\left(\frac{4n + 5}{2n + 5}\right)x\right|

    \displaystyle = \lim_{n \to \infty}\left|\left(\frac{4 + \frac{5}{n}}{2 + \frac{5}{n}}\right)x\right|

    \displaystyle = \left|\left(\frac{4}{2}\right)x\right|

    \displaystyle = \left|2x\right|.


    The series is convergent when this limit \displaystyle <1. So

    \displaystyle |2x| < 1

    \displaystyle -1 < 2x < 1

    \displaystyle -\frac{1}{2} < x < \frac{1}{2}.


    So the radius of convergence is \displaystyle \frac{1}{2}, not \displaystyle \frac{1}{6}...

    Now substitute each of the endpoints of this interval of convergence into your original series, and test the convergence of those series...
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  4. #4
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    Quote Originally Posted by Prove It View Post
    1. Your series is \displaystyle \sum_{n = 0}^{\infty}\left[\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n\right].

    Using the ratio test

    \displaystyle \lim_{n \to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)[4(n+1)+1]}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)[2(n+1)+3]}x^{n+1}}{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n}\right|

    \displaystyle = \lim_{n\to \infty}\left|\left[\frac{4(n+1)+1}{2(n+1)+3}\right]x\right|

    \displaystyle = \lim_{n \to \infty}\left|\left(\frac{4n + 5}{2n + 5}\right)x\right|

    \displaystyle = \lim_{n \to \infty}\left|\left(\frac{4 + \frac{5}{n}}{2 + \frac{5}{n}}\right)x\right|

    \displaystyle = \left|\left(\frac{4}{2}\right)x\right|

    \displaystyle = \left|2x\right|.


    The series is convergent when this limit \displaystyle <1. So

    \displaystyle |2x| < 1

    \displaystyle -1 < 2x < 1

    \displaystyle -\frac{1}{2} < x < \frac{1}{2}.


    So the radius of convergence is \displaystyle \frac{1}{2}, not \displaystyle \frac{1}{6}...

    Now substitute each of the endpoints of this interval of convergence into your original series, and test the convergence of those series...
    should we not just ignore the term x^n when we do the ration test, this is how I do it

    |\displaystyle \frac{a_{n+1}}{a_n}| = \displaystyle \frac{5.9.13.17...(4n+5)}{5.7.9.11...(2n+5)} \displaystyle \frac{3.5.7...(2n+3)}{1.5.9...(4n+1)}
    reduced to \displaystyle \frac{3(4n+5)}{2n+5}= \displaystyle \frac{12+15/n}{2+5/n} which approach to 6 as n goes to infinity, so R is 1/6
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  5. #5
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    Absolutely not! The changing \displaystyle x terms will change the value of the series.

    The whole point in using the ratio test is to find the values of \displaystyle x for which the series is going to converge. If you leave them out, then you're only determining the convergence of the series when \displaystyle x=1.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Absolutely not! The changing \displaystyle x terms will change the value of the series.

    The whole point in using the ratio test is to find the values of \displaystyle x for which the series is going to converge. If you leave them out, then you're only determining the convergence of the series when \displaystyle x=1.
    well. we do have a theorem: If lim_{n-->infinity} |\displaystyle \frac{a_{n+1}}{a_n}|=L>0, then \displaystyle \sum a_nx^n has a radius convergence R=1/L. If L=0, then R=infinity and if L-->infinity as n-->infinity then R=0.
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  7. #7
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    You're not listening to me.

    The \displaystyle x^n is part of \displaystyle a_n. If you're going to use the ratio test, then you need to use ALL of \displaystyle a_{n + 1} and \displaystyle a_n. In fact, the \displaystyle x term is the most important part, because when you have evaluated what \displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| is, you will get an expression for \displaystyle x. THIS IS THE IMPORTANT PART, because you know that the series will converge when this limit is \displaystyle <1, which means you can solve to find the values of \displaystyle x for which the series will converge. THIS abosolute value is the radius of convergence, and the resulting interval is the interval of convergence.
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  8. #8
    MHF Contributor matheagle's Avatar
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    i am so listening
    but I'm busy grading exams
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