# GRE 9367 Problem #38 (Summations + Limits)

1. Oct 28, 2008

### moo5003

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

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

ie:
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. Oct 28, 2008

### Dick

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.

3. Oct 29, 2008

### moo5003

Sorry you are correct, I didnt write it out correctly. I edited the first post so it should be correct now.

4. Oct 29, 2008

### Dick

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.

5. Oct 29, 2008

### moo5003

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?

6. Oct 29, 2008

### Dick

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.