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Math Help - Series convergent by AST, divergent by LCT?

  1. #1
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    Series convergent by AST, divergent by LCT?

    Hello,

    I have a question regarding the following problem:

    Describe if the following series converges absolutely, conditionally or diverges.

    Sigma (n=2 to infinity) (-1)^n * (n^2+2n)/(n^3-3n))

    Using the "absolute" ratio test, I get lim = 1, so no conclusion can be drawn.

    Using the "absolute" limit comparison test with 1/n, I get lim = 1, concluding that it diverges. (as 1/n diverges)

    Using the alternating series test this series a) is decreasing and b) the limit goes to 0; concluding that this series converges.

    As the absolute part of the series diverges by the limit comparison test, but the absolute part of the series converge by the alternating series test, can I conclude this series converges conditionally?

    Thanks
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  2. #2
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    Quote Originally Posted by FlyingDan View Post
    Hello,

    I have a question regarding the following problem:

    Describe if the following series converges absolutely, conditionally or diverges.

    Sigma (n=2 to infinity) (-1)^n * (n^2+2n)/(n^3-3n))

    Using the "absolute" ratio test, I get lim = 1, so no conclusion can be drawn.

    Using the "absolute" limit comparison test with 1/n, I get lim = 1, concluding that it diverges. (as 1/n diverges)

    Using the alternating series test this series a) is decreasing and b) the limit goes to 0; concluding that this series converges.

    As the absolute part of the series diverges by the limit comparison test, but the absolute part of the series converge by the alternating series test, can I conclude this series converges conditionally?

    Thanks
    Absolute convergence: Use the comparison test. Note that \frac{n^2+2}{n^3 - 3} > \frac{n^2}{n^3} = \frac{1}{n}.
    Last edited by mr fantastic; November 16th 2008 at 04:56 AM. Reason: Misreading but the argument and result is the same.
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  3. #3
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    Thanks for your reply, Mr Fantastic.

    Quote Originally Posted by mr fantastic View Post
    Absolute convergence: Use the comparison test. Note that \frac{n+2}{n^2 - 3} > \frac{n}{n^2} = \frac{1}{n}.
    Am I missing something here? The series I was referring to is \frac{n^2+2}{n^3 - 3}.
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  4. #4
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    Quote Originally Posted by FlyingDan View Post
    Thanks for your reply, Mr Fantastic.



    Am I missing something here? The series I was referring to is \frac{n^2+2}{n^3 - 3}.
    I misread the expression. I've edited my reply. The argument and result remain exactly the same.
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    Quote Originally Posted by mr fantastic View Post
    I misread the expression. I've edited my reply. The argument and result remain exactly the same.
    I'm still a bit lost though. Doesnt 1/n diverge as it is a p series with p=1?
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  6. #6
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    Quote Originally Posted by FlyingDan View Post
    Hello,

    I have a question regarding the following problem:

    Describe if the following series converges absolutely, conditionally or diverges.

    Sigma (n=2 to infinity) (-1)^n * (n^2+2n)/(n^3-3n))

    Using the "absolute" ratio test, I get lim = 1, so no conclusion can be drawn.

    Using the "absolute" limit comparison test with 1/n, I get lim = 1, concluding that it diverges. (as 1/n diverges)

    Using the alternating series test this series a) is decreasing and b) the limit goes to 0; concluding that this series converges.

    As the absolute part of the series diverges by the limit comparison test, but the absolute part of the series converge by the alternating series test, can I conclude this series converges conditionally?

    Thanks
    As Mr. Fantastic noted \sum_{n=1}^{\infty}\left|\frac{(-1)^n(n^2+2n)}{n^3-3}\right|=\sum_{n=1}^{\infty}\frac{n^2+2n}{n^3-3} and that last series diverges by limit comparison to the harmonci series. But note that if a_n=\frac{n^2+n}{n^3-3} that a_{n+1}\leqslant{a_n} and \lim_{n\to\infty}a_n=0, so \sum_{n=1}^{\infty}\frac{(-1)^n(n^2+2n)}{n^3-3} converges by the Leibniz's Criterion (AST). So we conclude that \sum_{n=1}^{\infty}\frac{(-1)^n(n^2+2n)}{n^3-3} converges conditionally.
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  7. #7
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    Quote Originally Posted by FlyingDan View Post
    I'm still a bit lost though. Doesnt 1/n diverge as it is a p series with p=1?
    Yes it does . And if you're familiar with the comparison test you will know that if u_n > v_n and \sum v_n diverges then \sum u_n diverges.
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