# Real Analysis: Riemann Sums

1. Oct 22, 2008

### steelphantom

1. The problem statement, all variables and given/known data
Find the limit, as n -> infinity, of $$\sum_{k=1}^n$$k3/n4

2. Relevant equations
Riemann sum: S(f, $$\pi$$, $$\sigma$$) = $$\sum_{k=1}^n$$f($$\xi$$)(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. Oct 23, 2008

### 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.

3. Oct 23, 2008

### Dick

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.

4. Oct 23, 2008

### steelphantom

As usual, I'm just not seeing it. 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 = $$\int$$x^3 from a to b?

By using the definition of a Riemann sum, I get: $$\sum_{k=1}^n \xi$$3(xk-xk-1). But what are the xks?

5. Oct 23, 2008

### Dick

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
6. Oct 23, 2008

### steelphantom

Ok, so if I choose $$\xi$$ = k/n = xk, then I get

$$\sum$$(k/n)3(k/n - (k-1)/n) = $$\sum$$(k3/n3)(1/n) = $$\sum$$k3/n4.

So the limit of this sum is $$\int_{0}^1$$x^3 ?

7. Oct 23, 2008

### Dick

Exactly.

8. Oct 23, 2008

### steelphantom

Cool. Thanks!

9. Oct 23, 2008

### steelphantom

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 $$\sum_{k=1}^n$$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 $$\xi$$ 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
10. Oct 23, 2008

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

can be written as a 4th degree polynomial in n: that will make your limit as $$n \to \infty$$ easier to evaluate

11. Oct 23, 2008

### steelphantom

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.

12. Oct 23, 2008

### Dick

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
13. Oct 23, 2008

### steelphantom

Got it. Thanks!