1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

GRE 9367 Problem #38 (Summations + Limits)

  1. Oct 28, 2008 #1
    1. The problem statement, all variables and given/known data

    The limit as n approaches infinity of:
    The sum from i = 1 to n of:

    (3/n)^3 * i^2 - (3/n)^2 * i

    A) -1/6
    B) 0
    C) 3
    D) 9/2
    E) 31/6

    Answer - D

    3. The attempt at a solution

    So, my first idea was to split the sum into two distinct sums and then take out powers of 3/n

    Limit of:

    27/n^3 Sum of i^2


    Limit of:

    9/n^2 Sum of -i

    Now the second sum is just -(n+1)n/2 thus the second term goes to -9/2 as n goes to infinity.

    This is were I'm stuck any help would be appreciated. Or suggestions on approaching this problem differently.
    Last edited: Oct 29, 2008
  2. jcsd
  3. Oct 28, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Why do you only have sum(i^2) in the first one. Isn't it sum(i^3)? Are you sure you've stated the problem correctly? As stated, the limit diverges.
  4. Oct 29, 2008 #3
    Sorry you are correct, I didnt write it out correctly. I edited the first post so it should be correct now.
  5. Oct 29, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    The sum of i^2 from 1 to n is n(n+1)(2n+1)/6. That's one way to do it. The easy way is to recognize it as a Riemann sum for the function 27x^2-9x integrated between 0 and 1.
  6. Oct 29, 2008 #5
    Thanks for the sum on i^2 seems to work out. I'm a little unclear how you recognize that the sum is infact a riemann sum integrated between 0 and 1.

    I can see where the function is namely 27i^2-9i however could you explain in further detail how the interval is seen?
  7. Oct 29, 2008 #6


    User Avatar
    Science Advisor
    Homework Helper

    Split the interval [0,1] into n equal intervals. The value of x at the end of an interval is i/n. The width of a rectangle is 1/n. The height of the rectangle corresponding to f(x)=27x^2-9x is 27(i/n)^2-9(i/n). Multiply height time width and that the contribution to the area sum from each rectangle. That's your sum.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: GRE 9367 Problem #38 (Summations + Limits)
  1. Limit of a Summation. (Replies: 6)

  2. Limit of summation (Replies: 2)