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Math Help - estimation of series sum

  1. #1
    Super Member 11rdc11's Avatar
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    estimation of series sum

    I'm having problems with this question

    Estimate \sum_{n=1}^{\infty}n^{-\frac{3}{2}} to within .01

    I know I can use the integral test with my remainder estimate being .01 but it would take such a large number to find the answer.

    -\frac{2}{\sqrt{x}} < .01

    I would have to sum up 40,000 terms to get it to that accuracy

    Is there another way to do this? Thanks in advance
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  2. #2
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    trying to think of an easier way, nothing is coming to mind... doesnt look like you can manipulate this to be a geometric series in any way.


    I will tell you that my old graphic calculator would do a summation for you, up to a certain bound...
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by damadama View Post
    trying to think of an easier way, nothing is coming to mind... doesnt look like you can manipulate this to be a geometric series in any way.


    I will tell you that my old graphic calculator would do a summation for you, up to a certain bound...
    Yea i tried using my graphing caculator, poor thing is still displaying busy
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  4. #4
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    You might be able to apply the Euler-Maclaurin Summation Formula:

    Euler?Maclaurin formula - Wikipedia, the free encyclopedia
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  5. #5
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    Quote Originally Posted by 11rdc11 View Post
    I'm having problems with this question

    Estimate \sum_{n=1}^{\infty}n^{-\frac{3}{2}} to within .01

    I know I can use the integral test with my remainder estimate being .01 but it would take such a large number to find the answer.

    -\frac{2}{\sqrt{x}} < .01

    I would have to sum up 40,000 terms to get it to that accuracy

    Is there another way to do this? Thanks in advance
    \sum_{n=1}^{\infty}n^{-\frac{3}{2}} \leq \int_{1}^{\infty}x^{-\frac{3}{2}}\,dx

    \int_{1}^{\infty}x^{-\frac{3}{2}}\,dx=\lim_{\varepsilon->\infty} \int_{1}^{\varepsilon}x^{-\frac{3}{2}}\,dx
    =\lim_{\varepsilon->\infty}\left(\frac{-2}{\sqrt{\varepsilon}}\right ) - \frac{-2}{\sqrt{1}}=2

    So \sum_{n=1}^{\infty}n^{-\frac{3}{2}} \leq 2

    I hope that's right... I haven't done integral approximations in a long time...
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  6. #6
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Prove It View Post
    \sum_{n=1}^{\infty}n^{-\frac{3}{2}} \leq \int_{1}^{\infty}x^{-\frac{3}{2}}\,dx

    \int_{1}^{\infty}x^{-\frac{3}{2}}\,dx=\lim_{\varepsilon->\infty} \int_{1}^{\varepsilon}x^{-\frac{3}{2}}\,dx
    Sorry im confused what do you mean?
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  7. #7
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    Quote Originally Posted by 11rdc11 View Post
    Sorry im confused what do you mean?
    OK, think back to elementary calculus. A definite integral is really an infinite sum of the trapezoids made under the curve. The sum of terms of n^{\frac{-3}{2}} therefore can not be any greater than an integral with the same limits. Evaluate the integral and you have an upper bound for the sum.
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