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Homework Help: Real Analysis: Riemann Sums

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the limit, as n -> infinity, of [tex]\sum_{k=1}^n[/tex]k3/n4

    2. Relevant equations
    Riemann sum: S(f, [tex]\pi[/tex], [tex]\sigma[/tex]) = [tex]\sum_{k=1}^n[/tex]f([tex]\xi[/tex])(xk - xk-1)

    3. The attempt at a solution
    My guess is that I should try to put this sum in terms of a Riemann sum, and then taking n -> infinity will give an integral of something. I'm just not sure what it is. Any hints? Thanks!
     
  2. jcsd
  3. Oct 23, 2008 #2

    Mark44

    Staff: Mentor

    I don't think you'll be successful with a Riemann sum.

    I tried a few approaches, and then started to look at the sequence of partial sums, S_n, where S_n = 1^3/n^4 + 2^3/n^4 + ... + n^3/n^4
    = 1/n^4 * (1^3 + 2^3 + ... + n^3)

    I started adding up the terms in the sum on the right and found that:
    1^3 + 2^3 = 9
    1^3 + 2^3 + 3^3 = 36
    1^3 + 2^3 + 3^3 + 4^3 = 100
    1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225

    There seemed to be a pattern here: each of the rightmost numbers is a perfect square, and further, each perfect square was directly related to the numbers being cubed on the left side of the equals sign.

    In other words,

    1^3 + 2^3 = 9 = 3^2
    1^3 + 2^3 + 3^3 = 36 = 6^2
    1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2
    1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225 = 15^2

    In the first bunch, you have 1 + 2 = 3
    In the second, you have 1 + 2 + 3 = 6
    In the third, you have 1 + 2 + 3 + 4 = 10
    In the fourth, you have 1 + 2 + 3 + 4 + 5 = 15

    I don't know if this is enough of a hint to get you all the way through this problem, but maybe it is. You'll still need to do some more work to find out whether S_n has a limit.
     
  4. Oct 23, 2008 #3

    Dick

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    It's pretty likely to be a Riemann sum for f(x)=x^3. Don't you agree? Try it and see if it works. It does.
     
  5. Oct 23, 2008 #4
    As usual, I'm just not seeing it. :redface: I guess the k^3 should be a dead giveaway, but I'm not sure what to do about the n^4. Is the limit = [tex]\int[/tex]x^3 from a to b?

    By using the definition of a Riemann sum, I get: [tex]\sum_{k=1}^n \xi[/tex]3(xk-xk-1). But what are the xks?
     
  6. Oct 23, 2008 #5

    Dick

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    Divide the interval [0,1] into n equal parts. The x_k's are the boundaries of the subintervals {1/n,2/n,3/n,...n/n}. Which point in each interval would be a good choice for 'xi'??
     
    Last edited: Oct 23, 2008
  7. Oct 23, 2008 #6
    Ok, so if I choose [tex]\xi[/tex] = k/n = xk, then I get

    [tex]\sum[/tex](k/n)3(k/n - (k-1)/n) = [tex]\sum[/tex](k3/n3)(1/n) = [tex]\sum[/tex]k3/n4.

    So the limit of this sum is [tex]\int_{0}^1[/tex]x^3 ?
     
  8. Oct 23, 2008 #7

    Dick

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    Exactly.
     
  9. Oct 23, 2008 #8
    Cool. Thanks!
     
  10. Oct 23, 2008 #9
    Ok, I have another Riemann sum, but this time, it actually says what the integral evaluates to. The question is the following:

    Show that the limit as n -> infinity [tex]\sum_{k=1}^n[/tex]n/(n2+k2) = pi/4.

    So I'm thinking that f(x) = sqrt(1-x2) on [0, 1]. But I'm having a hard time figuring out what to choose for [tex]\xi[/tex] and xk so that the sums match up. I wouldn't just say that xk = k/n again, would I? It doesn't seem like that would work.
     
    Last edited: Oct 23, 2008
  11. Oct 23, 2008 #10

    statdad

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    There is a closed-form expression that will help with this.
    [tex]
    \sum_{k=1}^n k^3
    [/tex]

    can be written as a 4th degree polynomial in n: that will make your limit as [tex] n \to \infty [/tex] easier to evaluate
     
  12. Oct 23, 2008 #11
    Thanks statdad, but I've already figured that one out with Dick's help. See the post right before yours for the one I'm now having trouble with.
     
  13. Oct 23, 2008 #12

    Dick

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    Since you got the last one, you know there should be a (1/n) for the interval length and the rest should be the f(k/n) part. So write it as (1/n)*(n^2/(n^2+k^2)). Do you see it yet? Can you write n^2/(n^2+k^2) as a function of k/n?
     
    Last edited: Oct 23, 2008
  14. Oct 23, 2008 #13
    Got it. Thanks!
     
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