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Math Help - Power Series Interval of Convergence

  1. #1
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    Power Series Interval of Convergence

    hi
    need help to see if i am doing this right.

    i have the series

    E (-1)^n+1 (x-1)^n / 5n

    i have to find the interval of convergence for f(x), f'(x), F"(x), and the intergral of f.

    i got the interval of convergence for f(x) but not sure its right. -9<x<11

    for f'(x) i have

    E (-1)^n+1 n(x-1)^n-1 / 5n

    which equals (1/5)(x-1)<1
    which equals -6<x<6

    well im not sure im doing this right so yeah any help?
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  2. #2
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    Quote Originally Posted by Ricky007 View Post
    hi
    need help to see if i am doing this right.

    i have the series

    E (-1)^n+1 (x-1)^n / 5n

    i have to find the interval of convergence for f(x), f'(x), F"(x), and the intergral of f.
    Are we supposed to guess that f(x) is that sum?

    i got the interval of convergence for f(x) but not sure its right. -9<x<11
    That is not what I get. How did you get that?

    for f'(x) i have

    E (-1)^n+1 n(x-1)^n-1 / 5n

    which equals (1/5)(x-1)<1
    which equals -6<x<6
    No, the sum is not equal to that! Perhaps you meant that the ratio test reduces to that- but that is not true either. How did you get (1/5)(x-1)< 1?

    well im not sure im doing this right so yeah any help?
    Pretty much nothing is correct here but since you don't show how you got these, I can't see how to correct it.
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  3. #3
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    help

    this is what i did. what did you get?

    interval of convergence for f

    E (-1)^n+1 (x-1)^n / 5n

    lim (u_n+1 / u_n) = lim [(-1)^n+2 (x-1)^n+1 / 5n+10][5n / (-1)^n+1 (x-1)^n]

    lim (-1)^n (x-1) / 10

    (x-1) / 10 <1

    -10<x-1<10
    -9<x<11

    im so lost here someone help!
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  4. #4
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    Quote Originally Posted by Ricky007 View Post
    this is what i did. what did you get?

    interval of convergence for f

    E (-1)^n+1 (x-1)^n / 5n

    lim (u_n+1 / u_n) = lim [(-1)^n+2 (x-1)^n+1 / 5n+10][5n / (-1)^n+1 (x-1)^n]
    First, since the ratio test only applies to positive series, that should be absolute value of the ratio so the -1 disappears immediately. You should have \frac{|x-1|^{n+1}}{5(n+1)}\frac{5n}{|x-1|^n= \frac{n+1}{n}|x-1|. Now the only question is "What is the limit of \frac{n+1}{n} as n goes to infinity?"

    lim (-1)^n (x-1) / 10

    (x-1) / 10 <1

    -10<x-1<10
    -9<x<11

    im so lost here someone help!
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