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Finite Sum Convergence

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data[/b]

    See attached

    2. Relevant equations

    See attached

    3. The attempt at a solution

    I know the answer is 6 or zero... but I cant figure out how to "show" this. When typing this equation into my calculator, I can clearly see that the number always ends in .0 or .6.

    Initially I let k = (n-(n-1))

    so the equation turns into

    (n-(n-1))*([n-(n-1)] +1) / n = n(n-1)/n (1)

    I then solve for n and I get

    n^2 + 3n - 2 = 0

    My calculator produces results

    n^2 + 3n + 2 = 0.

    Is equation (1) the correct way to solve this problem? If it is, the I just have a sign error somewhere... but I've triple checked my work and it seems there is none, so I am assuming (1) is not correct. Any suggestions?
     

    Attached Files:

  2. jcsd
  3. Nov 18, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    I've never seen this type of question before, so I wouldn't know how to solve it right away.
    But I do want to point out that n-(n-1)) = 1. So you are letting k = 1, is that what you meant to say?

    By the looks of it your hunch is right though, the sum is always either an integer, or something of the form (integer) + (2/3). You might want to try finding a closed form of the result (i.e. a general formula for all n) and then consider n modulo 3...
     
  4. Nov 18, 2009 #3

    Mark44

    Staff: Mentor

    In the sum below, you can move the n in the denominator outside the summation.
    [tex]\sum_{k = 1}^n \frac{k(k + 1)}{n}~=~\frac{1}{n}\sum_{k = 1}^n k(k + 1)[/tex]

    Now you can rewrite k(k + 1) as k2 + k, and split the single sum into two separate summations, not forgetting the factor of 1/n for each.

    Using the well-known representations for [itex]\sum{k^2}[/itex] and [itex]\sum{k}[/itex], you can get a nice, neat expression for your original sum.
     
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