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Math Help - Series

  1. #1
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    Series

    Find the radius of convergence and interval of convergence of the series:

    Sum of n(x-4)^n / (n^3+1) with n = 1 to infinity.

    can someone help me with this? Thanks!
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  2. #2
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    a_n  = \frac{n}{{n^3  + 1}}\left( {x - 4} \right)^n

    \left( {\sqrt[n]{{\left| {a_n } \right|}}} \right) = \left( {\sqrt[n]{{\frac{n}{{n^3  + 1}}}}\left| {x - 4} \right|} \right) \to \left| {x - 4} \right|<br />
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  3. #3
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    Hello, xfyz!

    Find the radius of convergence and interval of convergence of the series:

    . . \sum^{\infty}_{n=1}\frac{n(x -4)^n}{(n^3+1)}
    I'd use the Ratio Test . . .

    \left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{(n+1)(x-4)^{n+1}}{(n+1)^3+1} \cdot \frac{n^3+1}{n(x-4)^n}\right| \;= . \left|\frac{n+1}{n}\cdot\frac{n^3+1}{n^3+3n^2+3n + 2}\cdot\frac{(x-4)^{n+1}}{(x+4)^n}\right|


    Divide top and bottom of the second fraction by n^3

    . . \left|\left(1 + \frac{1}{n}\right)\left(\frac{1 + \frac{1}{n^3}}{1 + \frac{3}{n} + \frac{3}{n^2} + \frac{2}{n^3}}\right)(x-4)\right|


    Take the limit:

    . . \lim_{n\to\infty} \left|\left(1 + \frac{1}{n}\right)\left(\frac{1 + \frac{1}{n^3}}{1 + \frac{3}{n} + \frac{3}{n^2} + \frac{2}{n^3}}\right)(x-4)\right| \;=\;\left|(1 + 0)\left(\frac{1+0}{1+0+0+0}\right)(x-4)\right| . \;=\;|x-4|<br />



    We have: . |x - 4| \:< \:1\quad\Rightarrow\quad -1 \:<\:x-4\:<\:1\quad\Rightarrow\quad 3 \:<\:x\:<\:5


    . . The radius of convergence is: . R \,=\,1

    . . The interval of convergence: . (3,\,5)

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