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Math Help - series again!

  1. #1
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    Post series again!

    Hey! I need some help again with series. I hate this topic lol

    how would you do these problems

    1. Finde the range of values for which the following series is convergent or divergent:
    \sum_{n=1}^{\infty}\frac{(n+1)}{n^3}x^n

    2. Investigate the convergence of the series

    \frac{1}{1.2}+\frac{x}{2.3}+\frac{x^2}{3.4}+\frac{  x^3}{4.5}+...for x>0

    sorry to flood you with questions, its because I have a test very soon on this.

    and want to get a good mark

    thank you!
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  2. #2
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    Quote Originally Posted by Enita View Post
    Hey! I need some help again with series. I hate this topic lol

    how would you do these problems

    1. Finde the range of values for which the following series is convergent or divergent:
    \sum_{n=1}^{\infty}\frac{(n+1)}{n^3}x^n
    Ratio test,
    \frac{n+2}{(n+1)^3}\cdot \frac{n^3}{n+1}
    I will leave you to check that the limit is 1.
    Thus,
    |x|<1
    We have absolute convergence.
    |x|>2
    We have divergence.
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  3. #3
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    re:

    Thanks for your reply.

    I wanted to know say you had:

    1. \frac{1}{1}+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3  }+...

    How do you know that:
    U_n=\frac{2n-1}{2^{n-1}}

    How would you work out U_n.

    Thank you
    Last edited by Enita; February 5th 2007 at 10:03 AM.
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  4. #4
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    does it require a general formula?
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  5. #5
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    Hello, Enita!

    I wanted to know, say you had: . \frac{1}{1}+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3  }+ \cdots

    How do you know that: . U_n\:=\:\frac{2n-1}{2^{n-1}}

    How would you work out U_n?

    You do it by Inspection . . . and a bit of Experience.
    . . You're expected to "eyeball" it.


    The numerators are: 1,\,3,\,5,\,7,\,\cdots
    . . You recognize them as consecutive odd numbers.
    And you're supposed to know the general formula: . a_n \:=\:2n-1

    The denominators are: 1,\,2,\,2^2,\,2^3,\,\cdots
    . . which you should recognize as consecutive powers of 2.
    Since it begins with 1, the first denominator must be 2^o.
    . . So the denominators are: . 2^{n-1}

    Therefore: . U_n \:=\:\frac{2n-1}{2^{n-1}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Another example: . \frac{5}{9} + \frac{8}{27} + \frac{11}{81} + \frac{14}{243} + \frac{17}{729} + \cdots


    The numerators are: 5,\,8,\,11,\,14,\,17,\cdots
    They "go up by 3" . . . This tells us that the formula is: 3n + a

    The first term is 8.
    So when n = 1, we have: . 3(1) + a \:=\:8\quad\Rightarrow\quad a \,=\,5
    . . Hence, the formula for the numerator is: . \boxed{3n + 5}


    The denominators are: 9,\,27,\,81,\,243,\,729,\cdots
    . . They go up by a multiple of 3. .The formula is 3^k
    Since it begins with 9, the first denominator is 3^2.
    . . So the denominators are: . \boxed{3^{n+1}}

    Therefore, the general term is: . U_n \:=\:\frac{3n+5}{3^{n+1}}


    Hope this helps . . .

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