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Math Help - Infinite Series

  1. #1
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    Infinite Series

    I'm having trouble with the following problem:

    Use the integral test to find upper and lower bounds on the limit of the series.

     \sum_{k=1}^{\infty} \frac {1}{k^3}
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    Quote Originally Posted by larson View Post
    I'm having trouble with the following problem:

    Use the integral test to find upper and lower bounds on the limit of the series.

     \sum_{k=1}^{\infty} \frac {1}{k^3}
    Drawing a graph and shading rectangles that represent the sum helps:

    \int_1^{\infty} \frac{1}{x^3} \, dx < \sum_{k=1}^{\infty} \frac{1}{k^3} < 1 + \int_1^{\infty} \frac{1}{x^3} \, dx
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    Quote Originally Posted by mr fantastic View Post
    Drawing a graph and shading rectangles that represent the sum helps:

    \int_1^{\infty} \frac{1}{x^3} \, dx < \sum_{k=1}^{\infty} \frac{1}{k^3} < 1 + \int_1^{\infty} \frac{1}{x^3} \, dx
    I get what your saying, but how would I go about graphing this... just graph  \frac {1}{x^3} and then do like a trapezoid sum or something? If so, what does this even mean? I'm pretty confused in this whole chapter.
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    Quote Originally Posted by larson View Post
    I get what your saying, but how would I go about graphing this... just graph  \frac {1}{x^3} and then do like a trapezoid sum or something? If so, what does this even mean? I'm pretty confused in this whole chapter.
    The integral can be evaluated easily.
     \int \frac {1}{x^3} \, dx = -\frac{1}{2x^2} + C

    Thus

    \int_1^{\infty} \frac{1}{x^3} \, dx < \sum_{k=1}^{\infty} \frac{1}{k^3} < 1 + \int_1^{\infty} \frac{1}{x^3} \, dx

    \frac12 < \sum_{k=1}^{\infty} \frac{1}{k^3} < \frac32
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    Quote Originally Posted by Isomorphism View Post
    The integral can be evaluated easily.
     \int \frac {1}{x^3} \, dx = -\frac{1}{2x^2} + C

    Thus

    \int_1^{\infty} \frac{1}{x^3} \, dx < \sum_{k=1}^{\infty} \frac{1}{k^3} < 1 + \int_1^{\infty} \frac{1}{x^3} \, dx

    \frac12 < \sum_{k=1}^{\infty} \frac{1}{k^3} < \frac32
    I get everything but the very last part... how do you get the 1/2 and the 3/2 in the end?
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     \int \frac {1}{x^3} \, dx = -\frac{1}{2x^2} + C

    So, \displaystyle \int_{1}^{\infty} \frac {1}{x^3} \, dx = -\frac{1}{2x^2}  \mid^{\infty}_{1} = 0 - \left(-\frac{1}{2}\right) = \frac12
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  7. #7
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    Quote Originally Posted by larson View Post
    I get everything but the very last part... how do you get the 1/2 and the 3/2 in the end?
    Do you know how to evaluate improper integrals?
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    Quote Originally Posted by mr fantastic View Post
    Do you know how to evaluate improper integrals?
    Yes, I understand how to get the 1/2, but how do you find the upper bound (the 3/4)?
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  9. #9
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    Quote Originally Posted by larson View Post
    Yes, I understand how to get the 1/2, but how do you find the upper bound (the 3/4)?
    Please read the replies more carefully.

    The upper bound is 1 + \int_1^{\infty} \frac{1}{x^3} \, dx.

    And you've already been shown that the integral is equal to 1/2 ......
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    Quote Originally Posted by mr fantastic View Post
    Please read the replies more carefully.

    The upper bound is 1 + \int_1^{\infty} \frac{1}{x^3} \, dx.

    And you've already been shown that the integral is equal to 1/2 ......
    Wowww... okay, thanks, sorry for that...
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