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Math Help - Series: Reasoning Test

  1. #1
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    Series: Ratio Test

    What are the conditions of the test of reason does not apply to a series ?
    Last edited by Apprentice123; September 13th 2009 at 10:23 AM.
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  2. #2
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    I have no idea at all what you mean by "test of reason"!

    Is this a language problem? Do you mean "ratio test"?

    (Hey, makes sense- ratio- rational- reason!)

    The ratio test for convergence of a series of positive numbers, \sum_{n=0}^\infty a_n, is to look at the limit of the ration \frac{a_{n+1}}{a_n}. If that limit is less than 1, the series converges. If it is greater than 1, the series converges. If it equal to one, it does not tell us if the series converges or not.

    So one situation in which the ratio test does not work (I am not sure I would say "does not apply") is when that limit of the ratio is 1.

    On the face of it, it looks like the ratio test can only apply to series of positive numbers, but if you have a series of all negative numbers, you could factor out -1 and apply it to the new series with positive numbers which converges if and only if the original series does.

    If you have a mixture of positive and negative numbers, you cannot apply the ratio test- although you could take the absolute value and determine absolute convergence. If a series converges absolutely, then it converges but there exist many series which converge, but not absolutely.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    I have no idea at all what you mean by "test of reason"!

    Is this a language problem? Do you mean "ratio test"?

    (Hey, makes sense- ratio- rational- reason!)

    The ratio test for convergence of a series of positive numbers, \sum_{n=0}^\infty a_n, is to look at the limit of the ration \frac{a_{n+1}}{a_n}. If that limit is less than 1, the series converges. If it is greater than 1, the series converges. If it equal to one, it does not tell us if the series converges or not.

    So one situation in which the ratio test does not work (I am not sure I would say "does not apply") is when that limit of the ratio is 1.

    On the face of it, it looks like the ratio test can only apply to series of positive numbers, but if you have a series of all negative numbers, you could factor out -1 and apply it to the new series with positive numbers which converges if and only if the original series does.

    If you have a mixture of positive and negative numbers, you cannot apply the ratio test- although you could take the absolute value and determine absolute convergence. If a series converges absolutely, then it converges but there exist many series which converge, but not absolutely.

    Thank you. Sorry about the confusion with the language
    Problem:

    Verify that the ratio test can be applied to investigate the convergence of each series. If so apply it, if not explain why the test can not be applied

    1) \sum_{n=1}^{\infty} \frac{n}{3^n}

    My solution:
    \frac{a_{n+1}}{a_n} = \frac{(n+1)}{3^{n+1}} . \frac{3^n}{n} = \frac{n+1}{3n}

    \lim_{n\to{\infty}} \frac{n+1}{3n} = \frac{1}{3}
    Convergent
    It is correct ?

    2) \sum_{n=0}^{\infty} \frac{n^2+1}{n^3+1}
    How do I calculate this?
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  4. #4
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    Quote Originally Posted by Apprentice123 View Post
    Thank you. Sorry about the confusion with the language
    Problem:

    Verify that the ratio test can be applied to investigate the convergence of each series. If so apply it, if not explain why the test can not be applied

    1) \sum_{n=1}^{\infty} \frac{n}{3^n}

    My solution:
    \frac{a_{n+1}}{a_n} = \frac{(n+1)}{3^{n+1}} . \frac{3^n}{n} = \frac{n+1}{3n}

    \lim_{n\to{\infty}} \frac{n+1}{3n} = \frac{1}{3}
    Convergent
    It is correct ?

    2) \sum_{n=0}^{\infty} \frac{n^2+1}{n^3+1}
    How do I calculate this?
    Ratio test is inconclusive. Try a limit comparison with \sum_{n=1}^{\infty} \frac{1}{n}.
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  5. #5
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    Quote Originally Posted by Danny View Post
    Ratio test is inconclusive. Try a limit comparison with \sum_{n=1}^{\infty} \frac{1}{n}.
    1) Is correct ?

    2) I can not use the ratio test. Why ?
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  6. #6
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    Quote Originally Posted by Apprentice123 View Post
    1) Is correct ?

    2) I can not use the ratio test. Why ?
    The first is correct. For the second

     <br />
\lim_{n \to \infty} \frac{\frac{(n+1)^2+1}{(n+1)^3+1}}{\frac{n^2+1}{n^  3+1}} = 1.<br />

    To drawn conclusions from the ratio test, if

     <br />
\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L<br />

    then L < 1 converent and if L > 0 divergent. No conclusion for L = 1.
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  7. #7
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    Quote Originally Posted by Danny View Post
    The first is correct. For the second

     <br />
\lim_{n \to \infty} \frac{\frac{(n+1)^2+1}{(n+1)^3+1}}{\frac{n^2+1}{n^  3+1}} = 1.<br />

    To drawn conclusions from the ratio test, if

     <br />
\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L<br />

    then L < 1 converent and if L > 0 divergent. No conclusion for L = 1.
    I'm not getting simplicity is expression

    \frac{\frac{(n+1)^2+1}{(n+1)^3+1}}{\frac{n^2+1}{n^  3+1}}
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  8. #8
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    Quote Originally Posted by Apprentice123 View Post
    I'm not getting simplicity is expression

    \frac{\frac{(n+1)^2+1}{(n+1)^3+1}}{\frac{n^2+1}{n^  3+1}}
    <br />
\frac{\frac{(n+1)^2+1}{(n+1)^3+1}}{\frac{n^2+1}{n^  3+1}} = <br />
\frac{(n+1)^2+1}{n^2+1} \cdot \frac{n^3+1}{(n+1)^3+1} = \frac{n^2+2n + 2}{n^2+1} \cdot \frac{n^3+1}{n^3 + 3n^2 + 3n + 2} <br /> <br />
     <br />
= \frac{1+\frac{2}{n} + \frac{2}{n^2}}{1 + \frac{1}{n^2}} \cdot \frac{1 + \frac{1}{n^3}}{1 + \frac{3}{x} + \frac{3}{n^2} + \frac{2}{n^3}}<br />

    as n \to \infty everthing drops except for the 1's.
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  9. #9
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    Quote Originally Posted by Danny View Post
    <br />
\frac{\frac{(n+1)^2+1}{(n+1)^3+1}}{\frac{n^2+1}{n^  3+1}} = <br />
\frac{(n+1)^2+1}{n^2+1} \cdot \frac{n^3+1}{(n+1)^3+1} = \frac{n^2+2n + 2}{n^2+1} \cdot \frac{n^3+1}{n^3 + 3n^2 + 3n + 2} <br /> <br />
     <br />
= \frac{1+\frac{2}{n} + \frac{2}{n^2}}{1 + \frac{1}{n^2}} \cdot \frac{1 + \frac{1}{n^3}}{1 + \frac{3}{x} + \frac{3}{n^2} + \frac{2}{n^3}}<br />

    as n \to \infty everthing drops except for the 1's.

    Thank you very much.
    How can I calculate with the ratio test
    \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(n^2-1)}{(n^3-1)}
    I know calculate with Leibnitz, but not in ratio test
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  10. #10
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    Quote Originally Posted by Apprentice123 View Post
    Thank you very much.
    How can I calculate with the ratio test
    \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(n^2-1)}{(n^3-1)}
    I know calculate with Leibnitz, but not in ratio test
    You can't - not with this particular series.
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  11. #11
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    Quote Originally Posted by Danny View Post
    You can't - not with this particular series.
    I can only resolve with test leibnitz?
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  12. #12
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    Quote Originally Posted by Apprentice123 View Post
    I can only resolve with test leibnitz?
    Yes, also called the aternating series test.
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  13. #13
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    Quote Originally Posted by Danny View Post
    Yes, also called the aternating series test.
    Thank you
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