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Math Help - Power Series

  1. #1
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    Power Series

    I am finding the radius of convergence and interval of convergence.

    the sum of n=1 to infinity of x^n/(6^n*n^4)

    i used the ratio test on this and found that the limit goes to 1 and the radius of convergence is 6.

    when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by aaronb View Post
    I am finding the radius of convergence and interval of convergence.

    the sum of n=1 to infinity of x^n/(6^n*n^4)

    i used the ratio test on this and found that the limit goes to 1 and the radius of convergence is 6.

    when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.
    a_n=\frac{x^n}{n^4\cdot6^n}

    a_{n+1}=\frac{x^{n+1}}{(n+1)^4\cdot6^{n+1}}=\frac{  x}{6}\cdot\frac{x^n}{(x+1)^46^n } = 1\cdot \frac{x^n}{(x+1)^46^n } <\frac{x^n}{n^46^n}=a_n

    Remember that \frac{x}{6}= 1 at the egde of the radius of convergence
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  3. #3
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    root test is faster than ratio test.
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  4. #4
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    Hello, aaronb!

    Interesting . . . I've seen the Ratio Test described in several ways now.
    . . And a few of them never form a ratio.


    Find the radius of convergence and interval of convergence:
    . . \sum^{\infty}_{n=1} \frac{x^n}{6^nn^4}

    R \:=\:\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{x^{n+1}}{6^{n+1}(n+1)^4}\cdot\fra  c{6^nn^4}{x^n}\right| \;=\;\left|\frac{6^n}{6^{n_1}}\cdot\frac{n^4}{(n+1  )^4} \cdot\frac{x^{n+1}}{x^n}\right|

    . . = \;\left|\,\frac{1}{6}\left(\frac{n}{n+1}\right)^4 x\,\right| \;=\; \left|\,\frac{1}{6}\left(\frac{1}{1 + \frac{1}{n}}\right)^4x\,\right|


    Then: . \lim_{n\to\infty} \left|\,\frac{1}{6}\left(\frac{1}{1 + \frac{1}{n}}\right)^4x\,\right| \;=\;\left|\frac{1}{6}\cdot1^4\cdot x\,\right| \;=\;\left|\frac{x}{6}\right|

    So we have: . \left|\frac{x}{6}\right| \:< \:1 \quad\Rightarrow\quad -6 \:<\:x\:<\:6

    . . The radius of convergence is: . R \,=\,6



    At x = \text{-}6\!:\;\;\sum^{\infty}_{n=1}\frac{(\text{-}6)^n}{6^nn^4} \;=\;\sum^{\infty}_{n=1}\frac{(\text{-}1)^n}{n^4}\quad\hdots an alternating p-series which converges

    At x = 6\!:\;\;\sum^{\infty}_{n=1}\frac{6^n}{6^nn^4} \;=\;\sum^{\infty}_{n=1}\frac{1}{n^4} \quad\hdots a convergent p-series

    . . The interval of convergence is: . [-6,\:6]

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  5. #5
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    Quote Originally Posted by aaronb View Post
    I am finding the radius of convergence and interval of convergence.

    the sum of n=1 to infinity of x^n/(6^n*n^4)

    i used the ratio test on this and found that the limit goes to 1
    I don't know what you mean by that. Do you mean that the limit of the ratios is less than 1 for |x|< 6?
    and the radius of convergence is 6.
    Yes, that is correct.

    when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.
    When x= -6, this is
    \sum_{n= 1}^\infty \frac{(-1)^}{6^n n^4}
    \frac{\frac{1}{6^n n^4}{\frac{1}{6^{n+1}(n+1)^4}= 6\left(\frac{n+1}{n}\right)^4
    It should be clear that that is larger than 6 so the numerator is larger than the denominator- the sequence is decreasing.
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