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Math Help - Interval of convergence and ratio test

  1. #1
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    Interval of convergence and ratio test

    I was wondering if I could get some help on these two problems...

    Find the interval of convergence....

    \sum^{\infty}_{n=2} \frac{(x-7)^{n}}{(2 ^ {n} {n}^2 ln n)}

    And use the ratio test to find convergence or divergence....

    \sum^{\infty}_{n=1} \frac{(n!)(2n)!}{(3n)!}


    Thanks, I'd really appreciate it. Some of it is done, but I am unsure of what to do from what I've got.
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  2. #2
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    Hello, Rosey!

    Here's the second one . . .


    Use the ratio test to find convergence or divergence: . \sum^{\infty}_{n=1} \frac{(n!)(2n)!}{(3n)!}

    R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!(2n+2)!}{(3n+3)!} \cdot\frac{(3n)!}{(n!)(2n)!} \;=\;\frac{(n+1)!}{n!}\cdot\frac{(2n+2)!}{(2n)!}\c  dot\frac{(3n)!}{(3n+3)!}

    . . = \;\frac{n+1}{1}\cdot\frac{(2n+2)(2n+1)}{1}\cdot\fr  ac{1}{(3n+3)(3n+2)(3n+1)} \;= \;\frac{(n+1)(2n+2)(2n+1)}{(3n+3)(3n+2)(3n+1)}


    Divide top and bottom by n^3

    . . \frac{\left(\dfrac{n+1}{n}\right)\left(\dfrac{2n+2  }{n}\right)\left(\dfrac{2n+1}{n}\right)} {\left(\dfrac{3n+3}{n}\right)\left(\dfrac{3n+2}{n}  \right)\left(\dfrac{3n+1}{n}\right)} \;= . \frac{\left(1 + \dfrac{1}{n}\right)\left(2 + \dfrac{2}{n}\right)\left(2 + \dfrac{1}{n}\right)} {\left(3 + \dfrac{3}{n}\right)\left(3 + \dfrac{2}{n}\right)\left(3 + \dfrac{1}{n}\right)}


    Then: . \lim_{n\to\infty} \frac{\left(1 + \dfrac{1}{n}\right)\left(2 + \dfrac{2}{n}\right)\left(2 + \dfrac{1}{n}\right)} {\left(3 + \dfrac{3}{n}\right)\left(3 + \dfrac{2}{n}\right)\left(3 + \dfrac{1}{n}\right)} \;=\;\frac{(1+0)(2+0)(2+0)}{(3+0)(3+0)(3+0)} \;=\;\frac{4}{27}

    Since \lim_{n\to\infty} R \;=\;\frac{4}{27}\;<\;1, the series converges.

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  3. #3
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    [quote=Rosey;299537]
    Find the interval of convergence....

    \sum^{\infty}_{n=2} \frac{(x-7)^{n}}{(2 ^ {n} {n}^2 ln n)}

    Ratio test would break this guy


    lim_{n-->\infty}\frac{a_{n+1}}{a_n} \
    lim_{n-->\infty}\frac{(x-7)^{n+1}}{(2^{n+1})(n+1)^2(ln(n+1)} \ /  \frac{(x-7)^{n}}{(2^{n})(n)^2(ln(n)} \

    Simplifying we get ...


    lim_{n-->\infty}\frac{(x-7)n^2 lnn}{(2)(n+1)^2(ln(n+1))}  =  \frac{x-7}{2}\

    5<x<9.


    We test the end points by pluging 5 and 9 into the original question.

    Doing that for 5 we obtain.

    \sum^{\infty}_{n=2} \frac{(-2)^{n}}{(2 ^ {n} {n}^2 ln n)}

    Which converges by alternating series test.

    When we plug in 9 we get

    \sum^{\infty}_{n=2} \frac{1}{( {n}^2 ln n)}

    Which converges by comparison test

     \frac{1}{n^2lnn}   <   \frac{1}{n^2}  \

    So the interval of convergence is

    [5,9]

    Radius of convergence is 2
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  4. #4
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    Thank you BOTH! Fantastic. I realised my error.

    I thought you just replace (n) by (n + 1) everywhere rather than (2n + 2) or (3n + 3)

    The other one I tried the root test...dunno where I went wrong there. Thanks again!
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