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Thread: 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:
    $\displaystyle \sum_{n=1}^{\infty}\frac{(n+1)}{n^3}x^n$

    2. Investigate the convergence of the series

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

    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:
    $\displaystyle \sum_{n=1}^{\infty}\frac{(n+1)}{n^3}x^n$
    Ratio test,
    $\displaystyle \frac{n+2}{(n+1)^3}\cdot \frac{n^3}{n+1}$
    I will leave you to check that the limit is 1.
    Thus,
    $\displaystyle |x|<1$
    We have absolute convergence.
    $\displaystyle |x|>2$
    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. $\displaystyle \frac{1}{1}+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3 }+...$

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

    How would you work out $\displaystyle U_n$.

    Thank you
    Last edited by Enita; Feb 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: .$\displaystyle \frac{1}{1}+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3 }+ \cdots$

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

    How would you work out $\displaystyle U_n$?

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


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

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

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


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

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


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

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


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

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


    Hope this helps . . .

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