Finding the limit of the sum k³/n⁴ as n approaches infinity

[steelphantom]

Homework Statement

Find the limit, as n -> infinity, of $$\sum_{k=1}^n$$k³/n⁴

Homework Equations

Riemann sum: S(f, $$\pi$$, $$\sigma$$) = $$\sum_{k=1}^n$$f($$\xi$$)(x_k - x_k-1)

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!

 

[Mark44]

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.

 

[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.

 

[steelphantom]

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.

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(x_k-x_k-1). But what are the x_ks?

 

[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'??

 

[steelphantom]

Ok, so if I choose $$\xi$$ = k/n = x_k, 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 ?

 

[Dick]

Ok, so if I choose $$\xi$$ = k/n = x_k, 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 ?

Exactly.

 

[steelphantom]

Cool. Thanks!

 

[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 x_k so that the sums match up. I wouldn't just say that x_k = k/n again, would I? It doesn't seem like that would work.

 

[statdad]

Homework Statement

Find the limit, as n -> infinity, of $$\sum_{k=1}^n$$k³/n⁴

Homework Equations

Riemann sum: S(f, $$\pi$$, $$\sigma$$) = $$\sum_{k=1}^n$$f($$\xi$$)(x_k - x_k-1)

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!

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

 

[steelphantom]

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

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.

 

[Dick]

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 x_k so that the sums match up. I wouldn't just say that x_k = k/n again, would I? It doesn't seem like that would work.

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?

 

[steelphantom]

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?

Got it. Thanks!