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Math Help - estimate error of partial sum, terms too complex to integrate

  1. #1
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    estimate error of partial sum, terms too complex to integrate

    Hi all,

    The problem is to calculate this sum to 10 terms and estimate the error:

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

    I summed the first 10 terms, no problem.

    If I understand correctly, the error is less than

    \int_{10}^{\infty} \frac{n}{(n+1)3^n} dn

    but I can't figure out how to integrate that. I have the feeling that I can just use

    \int_{10}^{\infty} \frac{1}{3^n} dn

    to approximate my error, but it seems like I need to say more to justify this.. Or maybe I'm just missing the integration technique for the more complex form.

    Thanks again for any help!

    Brian
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  2. #2
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    1) That's no good. n/(n+1) < 1 Why would you want to make it even bigger? It's already too big, just because it is an integral value and not a discrete sum. Of course, that IS an upper bound, but you may wish to find a smaller one.

    2) Why would you not start at 11? You already used 10.

    3) There are various ways to find the value of an integral. Don't give up because you don't see the anti-derivative. Starting at 11, I get around 4.739 x 10^(-6)
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    1) That's no good. n/(n+1) < 1 Why would you want to make it even bigger? It's already too big, just because it is an integral value and not a discrete sum. Of course, that IS an upper bound, but you may wish to find a smaller one.
    I figured that for very large values of n that 3^n became far more significant than n/(n+1). As you say, it is an upper bound, so perhaps I could somehow show that \int_n^\infty \frac{n}{n+1}dn is relatively insignificant? [Edit: oh, but that diverges..hm..]


    Quote Originally Posted by TKHunny View Post
    2) Why would you not start at 11? You already used 10.
    I originally thought that too. I chose 10 because if R_n = s - s_n then

    \int_{n+1}^\infty f(x)dx \leq R_n \leq \int_n^\infty f(x)dx

    so

    R_{10} \leq \int_{10}^\infty f(x)dx

    I should also note that examples from my book do the same thing; if they sum the first 100 terms, they then take the integral from 100 to infinity to estimate the error.


    Quote Originally Posted by TKHunny View Post
    3) There are various ways to find the value of an integral. Don't give up because you don't see the anti-derivative. Starting at 11, I get around 4.739 x 10^(-6)
    I think I need to show some kind of anti-differentiaion so as to be "showing my work"; I don't think numerical methods of integration are acceptable in my class, but I could be wrong.

    Thank you for your input,

    Brian
    Last edited by buckeye1973; February 4th 2010 at 06:16 AM.
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  4. #4
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    I rethought the "11" thing on my way to work. Sorry about that. That first term is likely to be the one than ensures it is an upper bound.

    If you insist on evaluating integrals, I thought of something that might be useful. One doesn't encounter it very often.

    Integration by parts yields:

    \int_{10}^{\infty}{\frac{x}{x+1}\frac{1}{3^{x}}}\;  dx\;=\;\frac{10}{11}\cdot\frac{1}{3^{10}\cdot\log(  3)}\;+\;\int_{10}^{\infty}{\frac{3^{-x}}{\log(3)}\cdot\frac{1}{(x+1)^2}}\;dx

    We just dropped two orders of magnitude in the approximation. Unfortunately, the known portion may not be an upper bound - due to the fact that the known portion is INCREASED by the integral to reach the exact value.

    Never one to be discouraged by ambiguous results, let's do it again!

    =\;\frac{10}{11}\cdot\frac{1}{3^{10}\cdot\log(3)}\  ;+\;\frac{1}{3^{10}\cdot(\log(3))^{2}\cdot 11^{2}} - Q

    Where Q is an integral with value in the neighborhood of 1.556x10^{-8}. We gained another order of magnitude AND we know it's an upper bound!

    Of course, one might want to worry about convergence. I didn't, but then I'm under no pressure to get the right answer.

    For what it's worth.
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  5. #5
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    Quote Originally Posted by TKHunny View Post
    Integration by parts yields:

    \int_{10}^{\infty}{\frac{x}{x+1}\frac{1}{3^{x}}}\;  dx\;=\;\frac{10}{11}\cdot\frac{1}{3^{10}\cdot\log(  3)}\;+\;\int_{10}^{\infty}{\frac{3^{-x}}{\log(3)}\cdot\frac{1}{(x+1)^2}}\;dx

    [et cetera...]
    I'll have to sit down and process that one later! But I do get the idea of it.

    Thank you again!

    Brian
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