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Math Help - interval of convergence of power series

  1. #1
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    interval of convergence of power series

    heres the problem: http://img211.imageshack.us/img211/4962/93204887ca3.png

    i used the ratio test to get 4n+6 |x|

    normally i would just subtract or add whats inside the absolute value to each side of x. and then divide by the 6|x| i believe.

    however,
    i'm really confused by the 4n+6, how do i get the radius of convergence now?
    also, its possible this problem diverges.

    thanks for any help.
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  2. #2
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    Hello, rcmango!

    That function doesn't look right . . . \sum^{\infty}_{n=1}\frac{(2n!)}{n!}\,x^n
    We can cancel the n! and have: . \sum^{\infty}_{n=1} 2x^n

    So I suspect that it is: . \sum^{\infty}_{n=1}\frac{(2n)!}{n!}x^n


    The ratio is: . \frac{[2(n+1)]!x^{n+1}}{(n+1)!} \cdot\frac{n!}{(2n)!x^n} \;=\;\frac{[2(n+1)]!}{(2n)!}\cdot\frac{n!}{(n+1)!} \cdot\frac{x^{n+1}}{x^n}

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

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  3. #3
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    Okay, which brings me to my question again, its okay just to say it diverges at this point?

    what is going on with the 2n+6 is it because the 2n? and i don't need to show any interval?
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