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Sum of Series

  1. Dec 7, 2014 #1
    Can someone explain how to compute the sum of the following series?
    1/(n2 + (r-1)2) + 1/(n2 + (r)2) + 1/(n2 + (r+1)2) + ....1/(n2 + (n-1)2)
     
  2. jcsd
  3. Dec 7, 2014 #2
    Is anyone there?
     
  4. Dec 7, 2014 #3

    Stephen Tashi

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    Explain the context. Do you seek a closed form expression that is simpler to write than the sum itself? Do you need an approximation?

    I suggest we first try to find the summation::
    [itex] S_m = \sum_{k=1}^m {\frac{1}{1 + k^2}} [/itex].
    since the result might be used to derive the summation of the series you gave.
     
  5. Dec 7, 2014 #4
    I don't have an idea as to how to go about finding even the summation you gave. Can you give me a clue?
     
  6. Dec 7, 2014 #5

    Stephen Tashi

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    I can only give suggestions. ( I could give a "clue" if I knew how to sum it already.)

    Apply the section "Rational Functions" in the article: http://en.wikipedia.org/wiki/List_of_mathematical_series

    [itex] \sum_{k=1}^\infty \frac{1}{1+k^2} = -\frac{1}{2} + \frac{\pi}{2} coth(\pi) [/itex]

    So perhaps a substitution involving the hyperbolic cotangent would be useful in doing the finite sum. I'll keep thinking about it.
     
  7. Dec 7, 2014 #6
    Woah. I think this is beyond what my curriculum asks for. I am pretty sure that it doesn't involve trigonometric solutions as simplification of any series. I think this has something else to do with. Definitely not this way.
     
  8. Dec 7, 2014 #7

    Stephen Tashi

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    What has your curriculum covered? Is this in a chapter on mathematical induction?
     
  9. Dec 7, 2014 #8
    Mathematical Induction in my curriculum doesn't include sum of series etc. It just involves proving LHS = RHS or proving the given statement by using induction. This topic is done in sequences and series. But it just involves basic sums like - ∑n, ∑n2, ∑n3. And I don't know how to incorporate these 3 into the summation above to simplify it.
     
  10. Dec 7, 2014 #9

    Stephen Tashi

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    Did you quote the problem exactly? - or does the statement of the problem use summation notation?
     
  11. Dec 8, 2014 #10
    Its a part of the problem that I told you. The exact problem is-

    Consider a function f(x) = 1/(1+x2)
    Let αn = 1/n *([itex] \sum_{r=1}^n {f(\frac{r}{n}}))[/itex]
    and βn = 1/n *([itex] \sum_{r=0}^{n-1} {f(\frac{r}{n}})) [/itex] ; n ∈ N
    α = limn →∞n ) & β = limn →∞n )
    then αn/β - βn/α will always be? (Answer: A real number)
     
    Last edited: Dec 8, 2014
  12. Dec 8, 2014 #11

    Stephen Tashi

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    Is the sum supposed to be [itex] \sum_{r=0}^{n-1} f(\frac{r}{n}) [/itex] ? It begins at [itex] r = 0 [/itex] ?
     
  13. Dec 8, 2014 #12
    Yes. I didn't know how to write it up. Anyways, I have edited it now. :)
     
  14. Dec 8, 2014 #13

    Stephen Tashi

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    My guess is that you don't have to find a summation formula in order to work the problem. I'd start wtih an equation relating [itex] \alpha_n [/itex] to [itex] \beta_n [/itex]

    [itex] \alpha_n + \frac{1}{n} f(\frac{0}{n}) - \frac{1}{n}f(\frac{n}{n}) = \beta_n [/itex]
     
  15. Dec 8, 2014 #14
    How did you come about this equation? I can't figure it out?
     
  16. Dec 8, 2014 #15

    Stephen Tashi

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    Compare the terms in the sum for [itex] \alpha_n [/itex] with those in the sum for [itex] \beta_n [/itex]. The sum for [itex] \alpha_n [/itex] is missing the [itex] r= 0 [/itex] term that [itex] \beta_n [/itex] has. The sum for [itex] \alpha_n [/itex] has a term for [itex] r = n [/itex] that [itex] \beta_n [/itex] does not.
     
  17. Dec 8, 2014 #16
    Oh yeah. I got that now. But the solution that I have says-
    [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n)) [/itex]
    And comparing to the clue you gave me I get-
    [itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) [/itex]
    This leaves me confused as to what went wrong.
    Here's the solution that I have. It confuses me everytime I go through it.
     

    Attached Files:

    Last edited: Dec 8, 2014
  18. Dec 8, 2014 #17
    No:D
     
  19. Dec 8, 2014 #18

    Stephen Tashi

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    There are two different problems. The problem you stated is related to Riemann sums for [itex] \int_0^1 f(x) dx [/itex]. The problem in the solution is related to Riemann sums for [itex]\int_0^n f(x) dx [/itex].

    In the problem you stated, the definition of [itex] \alpha_n [/itex] is a Riemann sum using the altitude at the right hand point of the base of each rectangle. In the solution, [itex] \alpha_n [/itex] is defined to use the altitutde at the left hand point of the base of each rectangle.
     
  20. Dec 8, 2014 #19
    Woah. This is not even a part of my course. I think you went a bit too far with that. Could you please explain what you said more clearly? How does the limit n→∞ change the limits from 0 to n to 0 to 1 in the integral?
     
  21. Dec 8, 2014 #20

    Stephen Tashi

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    Have you studied definite integrals? Riemann sums?
     
  22. Dec 8, 2014 #21
    I have studied definite integrals. Is Riemann sums just another name of it?
     
  23. Dec 8, 2014 #22

    Stephen Tashi

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    Definite integrals are usually defined as the limit of an approximation process that divides he area under the graph of a function into rectangles and sums their areas. Such sums are the Riemann sums.
     
  24. Dec 8, 2014 #23
    Yup. I know that alright. But still I am not able to understand how the limits change from (0 to n) to (0 to 1)?
     
  25. Dec 8, 2014 #24

    Stephen Tashi

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    How they change between the two problems is just arbitrary notation. Why they change between [itex] \alpha_n [/itex] and [itex] \beta_n [/itex] is because the symbols [itex] \alpha_n, \beta_n [/itex] are used to denote two different Riemann sums.

    One sum computes the area of a rectangle with a given base by using the value of the function at the left endpoint of the base. The other sum computes the area of a rectangle with a given base by using the value of the function at the right endpoint. In the particular problem we have an integral from x = 0 to x = 1. The sum that begins with the term [itex] \frac{1}{n} f(\frac{0}{n}) [/itex] is using the left endpoint of the base as the height of the rectangle. That sum ends with the term [itex] \frac{1}{n} f(\frac{n-1}{n}) [/itex] because the last rectangle in the interval [itex] [0,1] [/itex] has base [itex] [\frac{n-1}{n}, \frac{n}{n} ] [/itex]. So when we we define limits for the summation using an index [itex]k [/itex] , the index goes from [itex]k = 0 [/itex] to [itex]k = n-1 [/itex].

    The other Riemann sum computes the are of a given base by using the value of the function at the right end of the base. You should be able to see why an index [itex]k [/itex] for that sum would go from [itex] k = 1 [/itex] to [itex] k = n [/itex].
     
  26. Dec 8, 2014 #25

    Stephen Tashi

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    If you mean the limits of integration, they don't change for x. I was incorrect to say that. The solution page you has introduces a third variable "[itex] r[/itex]" that is also marked on the x-axis and their picture seems to indicated that [itex] r [/itex] goes from zero to infinity.
     
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