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Summation as Integral

  1. Dec 22, 2008 #1
    I have been trying to solve Summation as Limit to Infinity type of questions but there are hardly a few examples I could find in my book
    I know the general method for [tex]\lim_{n \rightarrow \infty } \frac{1}{n}\Sigma_{r=A(x)}^{B(x)}f\frac{r}{n}[/tex] where r/n is replaced by x and 1/n by dx, the limits adjusted and integrated.
    However, i am unable to understand how to apply this method if the function is f(r) and not of f(r/n)
    Eg. [tex] t_r=\frac{r}{1-3r^2+r^4}, \Sigma_{r=1}^{n} t_r = ? [/tex]
    Could someone plz explain this method or point me to some resources regarding this.
    Thanks

    PS: Convergence/Divergence isnt a part of my syllabus, yet.
     
  2. jcsd
  3. Dec 22, 2008 #2
    HOld on a minute to see whether i am getting u right.

    You are saying basically that how would one find the following limit, by recognizing it as a rimann sum right?

    [tex]\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}[/tex]

    Is this correct?
     
  4. Dec 22, 2008 #3
    yes the expression is correct, sorry I couldnt do proper LaTEX.
    But i am unfamiliar with Riemann Sums, if thats means definite integral, then yes. As I said, I only know the method for f(r/n) , not f(r).
    Thanks
     
  5. Dec 22, 2008 #4

    tiny-tim

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    Hi f(x)! :smile:
    Surely it's ((B-A)/n)∑r=1n f(A + r(B-A)/n)? :confused:
     
  6. Dec 22, 2008 #5

    HallsofIvy

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    [tex]\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}[/tex]
    is just, by definition,
    [tex]\sum_{r=1}^\infty\frac{r}{1-3r^2+r^4}[/tex]

    It can't be written as an integral (unless you use the Riemann-Stieljes integral).
     
  7. Dec 23, 2008 #6
    Hello Sir
    This is a problem from limits exercises in my textbook. Since Integral isnt possible, am I just supposed to find the sum using algebraic summation techniques and then limit as n-> infinity ?
    Could you plz give a hint how to find the sum ?
    Thanks

    Hello Sir
    The standard form I have in my text is the one I have put in my first post, but the one you have posted seems related (i think yours is the one with integral as sum of parts)
     
  8. Dec 23, 2008 #7

    tiny-tim

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    Hi f(x)! :smile:

    You could try partial fractions, and then integrating …

    though after that, I get stuck :redface:

    (and don't call us "Sir"!)
     
  9. Dec 23, 2008 #8

    Dick

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    Now why are YOU getting stuck, tiny-tim? That's a great suggestion. Now just complete the squares. One term contains (r+1/2)^2, the other (r-1/2)^2. It telescopes. Doesn't it?
     
  10. Dec 23, 2008 #9

    tiny-tim

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    ah! … got it! :biggrin:

    i didn't see that √[(3 + √5)/2] = (1 + √5)/2 :redface:

    EDIT: ooh, i didn't need to integrate either …

    as soon as you get the right 1/quadratic - 1/quadratic,

    you use n2 ± n = n(n ± 1),

    and the "telescoping" works immediately. :biggrin:
     
    Last edited: Dec 24, 2008
  11. Dec 23, 2008 #10

    hi, could you plz explain a little about how to write it as an Riemann-Stieljes integral? I learned something about Riemann-Stieljes integral in principle of mathematics but havn't met any concrete examples. Thanks
     
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