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Math Help - seriess

  1. #1
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    seriess

    hello everyonee!

    i need help on these questions please.


    1) determine wether each of the followin series is convergent:

    a) 2/2^2 + 2^2/2^2 + 2^3/3^2 + ... 2^n/n^2

    b) Un = 1+2n^2/1+n^2

    2)

    does this series in picture converge (sorry did not know how to write in text)
    Attached Thumbnails Attached Thumbnails seriess-untitled.gif  
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  2. #2
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    Quote Originally Posted by Enita View Post
    a) 2/2^2 + 2^2/2^2 + 2^3/3^2 + ... 2^n/n^2
    The general form is,
    \sum_{n=1}^{\infty} \frac{2^n}{n^2}
    Use the ratio test,
    \left| \frac{2^{n+1}}{(n+1)^2} \cdot \frac{n^2}{2^n} \right| =\left| \frac{2n^2}{n^2+2n+1}\right|=\left| \frac{2}{1+\frac{2}{n}+\frac{1}{n^2}}\right|
    The limit is 2 as n\to \infty.
    Thus, the ratio test says the series diverges.
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  3. #3
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    can you explain the ratio test method please?

    what the last line mean? starting with C
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  4. #4
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    Quote Originally Posted by Enita View Post

    b) Un = 1+2n^2/1+n^2
    The limits as n\to \infty is not equal to zero.
    Thus it diverges.
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  5. #5
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    Quote Originally Posted by Enita View Post
    can you explain the ratio test method please?
    I took,
    a_{n+1}=\frac{2^{n+1}}{(n+1)^2}
    And
    a_n=\frac{2^n}{n^2}
    Then divided them, (flip fractions)
    And also used the fact that,
    2^{n+1}=2\cdot 2^n.

    what the last line mean? starting with C
    That is the factor ring formed.... ignore that. It is my signature.
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  6. #6
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    oh ok

    i don't get this bit:
    =\left| \frac{2n^2}{n^2+2n+1}\right|=\left| \frac{2}{1+\frac{2}{n}+\frac{1}{n^2}}\right|
    The limit is 2 as n\to \infty.

    what if the limit was different?
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  7. #7
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    Quote Originally Posted by Enita View Post
    oh ok

    i don't get this bit:
    =\left| \frac{2n^2}{n^2+2n+1}\right|=\left| \frac{2}{1+\frac{2}{n}+\frac{1}{n^2}}\right|
    The limit is 2 as n\to \infty.

    what if the limit was different?
    Divide the numerator and denominator by n^2.
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  8. #8
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    hey thanks alot!

    I been working on these today and came across a example on the internet. But there is a process that i don't understand. I have attached a pic from the site and labelled what i don't understand with a red arrow. Hopefully if someone could help i would be very greatful.

    how does it cancel?

    (site: Calculus II (Math 2414) - Series & Sequences - Ratio Test)

    thanxs
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  9. #9
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    Okay,
    He expresses,
    4^{2n+1}=4^{2n}\cdot 4 in the numerator.
    And,
    4^{2n+3}=4^{2n}\cdot 4^3 in the denominator.
    After some canceling he gets,
    \frac{4}{4^3}=\frac{1}{4^2}
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  10. #10
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    Hello, Enita!

    Let's work with the fraction
    . . and ignore the limit and absolute value for now.


    We have: . \frac{(-10)^{n+1}} {4^{2n+3}(n+2)} \cdot\frac{4^{2n+1}(n + 1)}{(-10)^n} \;=\;\frac{(-10)^{n+1}}{(-10)^n} \cdot \frac{4^{2n+1}}{4^{2n+3}} \cdot\frac{n+1}{n+2}

    Simplify: . \frac{-10}{1}\cdot\frac{1}{4^2}\cdot\frac{n+1}{n+2} \;=\;\frac{-10(n+1)}{16(n+2)} . . . see?

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  11. #11
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    thanks. you guys are great! just like having a teacher right here.

    Thank you!
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  12. #12
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    Quote Originally Posted by Enita View Post
    thanks. you guys are great! just like having a teacher right here.
    No we are better we do it for free.
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  13. #13
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    does anyone know how to solve question 2?

    thanks math forum
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  14. #14
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    Quote Originally Posted by Enita View Post
    does anyone know how to solve question 2?

    thanks math forum
    That one is simple it is bounded by,
    a_n=\frac{1}{n!}
    And \sum a_n is known to converge.
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  15. #15
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    re:

    I worked this out for Un = \frac{1 + 2n^2}{1+n^2}


    \left| \frac{1+2(n+1)^2}{1+(n+1)^2} \cdot \frac{1+n^2}{1+2n^2} \right| = \left| \frac{1+2(n^2+2n+1)}{1+(n^2+2n+1)} \cdot \frac{1+n^2}{1+2n^2} \right|

    =\left| \frac{1+2n^2+4n+2}{1+n^2+2n+1} \cdot \frac{1+n^2}{1+2n^2} \right|=\left| \frac{4n+2}{2n+1}\right|

    divde by n
    =\left| \frac{4+2}{2+1}\right|<br />

    Series equals 6/3 which is > 1 so diverges?

    Have i worked this out right? (using the ratio test and canceling?)

    Thank you maths forum.
    Last edited by Enita; January 27th 2007 at 04:59 AM.
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