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What is the limit of 1/n^2 + 2/n^2 + . + n-1/n^2 as n-> infinity

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data

    find:

    lim(n[tex]\rightarrow\infty[/tex] (1/n^2 + 2/n^2 + 3/n^2 + ... + n-1/n^2 )

    2. Relevant equations



    3. The attempt at a solution[/b

    I could guess that the limit is zero but i dont know howto prove it
     
  2. jcsd
  3. Oct 25, 2008 #2

    HallsofIvy

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    Re: limit


    No, the limit is not 0.

    That sum is the same as (1/n2)(1+ 2+ 3+ ...+ (n-1)). Can you write that last sum in closed form?
     
  4. Oct 25, 2008 #3
    Re: limit

    Oh right,
    so, this is the same as:

    [tex]\frac{1}{n^2}[/tex][tex]\sumk[/tex] (from k=1 to n-1)

    and now i use the formula for geometric series?
     
  5. Oct 25, 2008 #4

    Mark44

    Staff: Mentor

    Re: limit

    Dick asked whether you knew a formula for 1 + 2 + 3 + ... + (n - 1). This is not a geometric series.
     
  6. Oct 25, 2008 #5
    Re: limit

    What is meant by 'closed form' ? i dont know, i thought dick meant that i should put it in a summation.
    i meant arithmetic...sorry, surely this is not geometric.
     
  7. Oct 25, 2008 #6
    Re: limit

    I have done it and i just want to make sure of my answer:
    i used the summation formula for aritmetic series :1/2 n(n+1)
    and got that the limit is 1/2
    is that right?
     
  8. Oct 25, 2008 #7

    Dick

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    Re: limit

    Sure it's right. You could also look at the problem as being a Riemann sum for the integral of f(x)=x from x=0 to x=1. Surely, 1/2. BTW, I didn't say anything before in this thread. That was Halls. Let's give credit where credit is due.
     
  9. Oct 25, 2008 #8

    Mark44

    Staff: Mentor

    Re: limit

    Sorry for giving credit where not due. I think I saw your name in another thread that sara started, and mistakenly cited you and Dirk.
     
  10. Oct 25, 2008 #9

    Dick

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    Re: limit

    No problem. If Hall's suggestion had been wrong, I would have been OUTRAGED to have it attributed to me. But it wasn't. :)
     
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