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Math Help - power series, radius of convergence.

  1. #1
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    power series, radius of convergence.

    I've posted a few steps of the problem i solved in the pic: http://img180.imageshack.us/img180/9913/26377643by1.png


    however, what is the radius of convergence? i got the interval to be [0, 1]

    do i just solve for x for the radius? |(2x-1)|

    thanks.
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  2. #2
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    Hello, rcmango!

    A small error in your algebra ... but you still got the right answer!


    \sum^{\infty}_{n=1}\frac{(2x-1)^n}{n^4+16}

    R \;=\;\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{(2x-1)^{n+1}}{(n+1)^4+16}\cdot\frac{n^4+16}{(2x-1)^n}\right| \;=\; \left|(2x-1)\cdot\frac{n^4+16}{(n+1)^4+16}\right|

    Then: . \lim_{n\to\infty} R \;=\;\lim_{n\to\infty}\left|(2x-1)\cdot\frac{n^4+16}{(n+1)^4+16}\right| \;=\;{\color{blue}|2x-1|}


    Hence: . |2x-1| \:<\:1\quad\Rightarrow\quad-1 \:<\:2x-1 \:<\:1\quad\Rightarrow\quad 0 \:<\:2x \:<\:2


    . . The interval of convergence is: . 0 \:<\:x \:<\:1

    . . And the radius of convergence is: . r \:=\:\frac{1}{2}

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  3. #3
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    btw, the 0 and the 1 those both converge then correct? so there both on the closed interval [0, 1] right?

    also, n^4 / (n+1)^4 does that just cancel?

    thanks alot.
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