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anonymous

  • 5 years ago

in order to show the function "[(lnx)/(x^3)]" either converges or diverges by the integral test, do i just integrate the function and evaluate?

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  1. anonymous
    • 5 years ago
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    Integrate the function from 1 to infinity -- if the integral converges to a specific value, then the series does. (That's the basic idea of the integral test.)

  2. anonymous
    • 5 years ago
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    got it.. thank you.. but i also have another question now... the book, it suggests that the given converges when compared to 1/(n^2)... why are they comparing to n^2 not n^3?

  3. anonymous
    • 5 years ago
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    It's just for simplicity's sake; they could have picked 1/n^a with an 'a' greater than 1. Correct me if I'm wrong, but that's the direct comparison test, right?

  4. anonymous
    • 5 years ago
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    they have similar one to this problem with [(lnn)/(n^(3/2))] and they did "comparing the series to a convergent p-series"... Limit Comparison Test?

  5. anonymous
    • 5 years ago
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    Yeah, they're basically saying that: 1.) If the function you want to evaluate is always less than 1/x^p (on the interval), and 1/x^p converges, then the function converges. 2.) If the function you want to evaluate is always less than 1/x^p (on the interval), and 1/x^p diverges, then the function diverges. When you evaluate 1/x^p and p>1, it converges. If p<1, it diverges.

  6. anonymous
    • 5 years ago
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    so i can simply just look at the exponent of the given denominator and decide on whether it converges or diverges?

  7. anonymous
    • 5 years ago
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    Not exactly, you can do that for a given function 1/x^p and see whether p is greater than or less than 1. If the function you're evaluating is ALWAYS less than a 1/x^p function on the interval, and you know if 1/x^p diverges or converges, then you can see if that function converges or diverges.

  8. anonymous
    • 5 years ago
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    thx x]

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