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Math Help - Ad hoc summation of series

  1. #16
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    When is it ok to interchange integration with summation to infinity?
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  2. #17
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Heirot View Post
    When is it ok to interchange integration with summation to infinity?
    If the partial sums of the summation converge uniformly.

    In your case, a way to test for uniform convergence is the Weierstrass M test.
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  3. #18
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    Thanks to all of you. I had a sum a while ago which to this day I cannot perform: 1/((n+1)(2n+1)(3n+1)), n=0,...,infinity. I tried using the trick of interchanging the summation with integration there and it didn't work. I'm wondering, is there another way of doing this sum?
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  4. #19
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Heirot View Post
    Thanks to all of you. I had a sum a while ago which to this day I cannot perform: 1/((n+1)(2n+1)(3n+1)), n=0,...,infinity. I tried using the trick of interchanging the summation with integration there and it didn't work. I'm wondering, is there another way of doing this sum?
     \displaystyle \frac1{(n+1)(2n+1)(3n+1)} = \frac a{n+1}+\frac b{2n+1}+\frac c{3n+1}

    So  \displaystyle \sum_{n=0}^\infty \frac1{(n+1)(2n+1)(3n+1)} = \int_0^1 \sum_{n=0}^\infty \left(ax^n+bx^{2n}+cx^{3n}\right) dx = \int_0^1 \left(\frac{a}{1-x}+\frac{b}{1-x^2}+\frac{c}{1-x^3}\right) dx for  |x|<1 .

    Can you take it from here?
    Last edited by chiph588@; January 2nd 2011 at 03:13 PM.
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  5. #20
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    I tried that - it doesn't work because the series don't converge uniformly so the integration and summation don't commute.
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  6. #21
    MHF Contributor chiph588@'s Avatar
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  7. #22
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    My previous thread about this subject: http://www.mathhelpforum.com/math-he...ce-156957.html

    Mathematica gives two different numerical answers.
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  8. #23
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    Quote Originally Posted by Soroban View Post
    Hello, Heirot!


    The general term is: . a_n \;=\;\dfrac{1}{n(n+1)(n+2)(n+3)}


    Partial fractions: . \displaystyle \frac{1}{n(n+1)(n+2)(n+3)} \;=\;\frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} + \frac{D}{n+3}

    We find that: . A = \frac{1}{6},\;B = -\frac{1}{2},\;C = \frac{1}{2},\;D = -\frac{1}{6}

    Hence: . \displaystyle a_n \;=\;\frac{\frac{1}{6}}{n} - \frac{\frac{1}{2}}{n+1} + \frac{\frac{1}{2}}{n+2} - \frac{\frac{1}{6}}{n+3} \;=\;\frac{1}{6}\left(\frac{1}{n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)

    Then: . \displaystyle S \;=\;\frac{1}{6}\sum^{\infty}_{n=1}\left(\frac{1}{  n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)


    Crank out several terms:

    \displaystyle S \;=\;\frac{1}{6}\bigg[\left(1 - \frac{3}{2} + \frac{3}{3} - \frac{1}{4}\right) + \left(\frac{1}{2} - \frac{3}{3} + \frac{3}{4} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6}\right)

    . . . . . . . \displaystyle + \left(\frac{1}{4} - \frac{3}{5} + \frac{3}{6} - \frac{1}{7}\right) + \left(\frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8}\right) + \left(\frac{1}{6} - \frac{3}{7} + \frac{3}{8} - \frac{1}{9}\right)

    . . . . . . . . . \displaystyle + \left(\frac{1}{7} - \frac{3}{8} + \frac{3}{9} - \frac{1}{10}\right) + \left(\frac{1}{8} - \frac{3}{9} + \frac{3}{10} - \frac{1}{11}\right) + \hdots \bigg]


    We find that nearly all the terms cancel out.

    And we are left with: . \displaystyle S \;=\;\frac{1}{6}\left(1 - \frac{3}{2} + \frac{1}{2} + \frac{1}{3}\right) \;=\;\frac{1}{18}

    I think I got you turned on to partial fractions (it's a wonderful method).
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  9. #24
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    Hello,

    I have another summation problem: 1/(1+a) + 1/((1+a)(1+2a)) + 1/((1+a)(1+2a)(1+3a)) + ...

    I'd like to express it as an integral, how should I proceed?
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  10. #25
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Heirot View Post
    Hello,

    I have another summation problem: 1/(1+a) + 1/((1+a)(1+2a)) + 1/((1+a)(1+2a)(1+3a)) + ...

    I'd like to express it as an integral, how should I proceed?
    Well \displaystyle\frac{1}{(1+a)\cdot (1+2a)...\cdot (1+k\cdot a)} = \frac{\left(\tfrac{1}{a}\right)^k}{\left(\tfrac{1}  {a}+1\right)\cdot ... \cdot \left(\tfrac{1}{a}+k\right)}

    Now note that we have \displaystyle\frac{1}{\left(\tfrac{1}{a}+1\right)\  cdot ... \cdot \left(\tfrac{1}{a}+k\right)} = \frac{\Gamma \left( \tfrac{1}{a} + 1\right)}{\Gamma(\tfrac{1}{a} + k + 1) } = \frac{\text{B}\left(\tfrac{1}{a}+1,k\right)}{\Gamm  a( k)}

    Try to go on from there.
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  11. #26
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    Comment and a question for you

    Quote Originally Posted by Heirot View Post
    Hello,

    I have another summation problem: 1/(1+a) + 1/((1+a)(1+2a)) + 1/((1+a)(1+2a)(1+3a)) + ...

    I'd like to express it as an integral, how should I proceed?
    My comment is that normally the integral is given first from which a series is derived. Here you have it in reverse which makes the problem interesting.

    My question is how do you know that the integral exists?
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  12. #27
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    My approach to complicated sums is to express them as integrals. Being a physicist, I find it easier to think in terms of integrals than sums. You can always transform a number into a definite integral. The question is to choose the simplest integral. This is where I sometimes need help from you guys
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