Convergence of Riemann Sums for Limit of Series

[springo]

Homework Statement

I need to find the following:
\lim_{n\rightarrow\infty}\left(\frac{1^2+2^2+3^2+...+n^2}{n^3} \right)

Homework Equations

The Attempt at a Solution

I know I could do the sum of the series to find the result but I would like to use Riemann sums.
I think I have to start by writing the limit as:
\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{n}\left(\frac{k}{n} \right)^2
However, I'm not sure how to continue.

Thanks for your help.

 

[snipez90]

1² + 2² + 3² + ... + n² = [n(n+1)(2n+1)]/6

 

[quasar987]

The thing with Riemann sums is that they are related to the Riemann integral through the equation

\lim_{\max\{\Delta x_i\}\rightarrow 0}\sum_{i=1}^n f(x_i^*)\Delta x_i=\int_a^bf(x)dx

This means that for a bounded integrable function f:[a,b]-->R and any sequence \{\mathcal{P}_n\}_{n\in\mathbb{N}} of partitions of [a,b] such that the largest subinterval in \mathcal{P}_n=\{a=x_0^{(n)}, x_1^{(n)},\ldots,x_N_n^{(n)}=b\} goes to zero as n\rightarrow\infty, the sequence of associated Riemann sums approaches the integral of f over [a,b]:

\left(\lim_{n\rightarrow\infty}\left[\max_{1\leq i\leq N_n}(\Delta x_i^{(n)})\right]= 0 \right)\Longrightarrow \left(\lim_{n\rightarrow\infty}\sum_{i=1}^{N_n}f(x_i^*)\Delta x_i^{(n)} = \int_a^bf(x)dx\right)

So...

What we have here is the limit

\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{n} \left(\frac{i}{n} \right)^2

Notice that for each positive integer n, {0=0/n, 1/n, 2/n ...,(n-1)/n, n/n=1} is a partition of the interval [0,1] and in such a partition, each subinterval has length 1/n. And 1/n\rightarrow 0 as n\rightarrow\infty, so according to the big implication above,

\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_i^*)\frac{1}{n} = \int_0^1f(x)dx

for any bounded integrable function f:[0,1]-->R.

Now, for which function f and choice of x_i^*\in [(i-1)/n,i/n] do we have

f(x_i^*)=\left(\frac{i}{n}\right)^2

??

Once you figure that out, you can use the fundamental theorem of calculus to find the value of the integral of f and hence the value of the sum you're interested in.

 

[springo]

OK, so if I understood well x ∈ [0,1] since 1/n is creating partitions of that interval.

Then we have x_i* ∈ [x_i-1, x_i].
Since x_i - x_i-1 = i/n we have x_i* = i/n.
Therefore f(x) = x².

Then:
\lim_{n\rightarrow\infty}\left(\frac{1^2+2^2+3^2+. ..+n^2}{n^3} \right) = \int_0^1 x^2 dx = \frac{1}{3}
I checked the result and it's correct, what about the proof?

Thanks again.

 

[quasar987]

OK, so if I understood well x ∈ [0,1] since 1/n is creating partitions of that interval.

Then we have x_i* ∈ [x_i-1, x_i].
Since x_i - x_i-1 = i/n we have x_i* = i/n.
Therefore f(x) = x².

Then:
\lim_{n\rightarrow\infty}\left(\frac{1^2+2^2+3^2+. ..+n^2}{n^3} \right) = \int_0^1 x^2 dx = \frac{1}{3}
I checked the result and it's correct, what about the proof?

Thanks again.

I'm not sure I agree with your chain of implications from a logical standpoint but you're the one who's best placed to say if you understand or not.

And what do you mean by "what about the proof"?

In any case, the way I would have solved the problem is simply by noting that, in view of what has been say above, the sums

\sum_{i=1}^{n}\frac{1}{n} \left(\frac{i}{n} \right)^2

are Riemann sums of the function f:[0,1]-->R defined by f(x)=x² where the points x_i^*\in [x_{i-1},x_i]=[(i-1)/n,i/n] have been chosen equal to i/n.

Hence,

<br /> \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{n} \left(\frac{i}{n} \right)^2 =\int_0^1x^2dx=1/3

 

[springo]

By the "proof" I meant the fact that the function is f(x) = x². But now I realize I made a mistake since x_i - x_i-1 = 1/n ≠ i/n.
So I don't understand how do you know the function is f(x) = x²?
Is it because x_i* = i/n, and if so, how can you find that?
Thanks.

 

[quasar987]

You do not know a priori that xi* = i/n.

Suppose more generally that you are given the sum following Riemann sum

\sum_{i=1}^{n}\frac{1}{n}F(i,n) \right)

where F(i,n) is some expression involving i and n.

The question is,

"For which function f:[0,1]-->R and which choice of points x_i^*\in [(i-1)/n,i/n] do we have f(x_i^*)=F(i,n) for all i=0,...,n ??"

In the present case, F(i,n)=(i/n)² and so after a moment of reflection trying various combinations of functions and points x_i^*, you will conclude that f(x)=x² and x_i^*=i/n does the trick.

 

[springo]

Thanks, I think I understood.

So would this be correct so far?
\lim_{n\rightarrow\infty}\left(\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+3n}}+...+\frac{1}{\sqrt{n^2+nn}} \right) = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{\sqrt{n}}\frac{1}{\sqrt{n+i}} = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{\sqrt{n}}f(x_i^*)=\int_{0}^{\infty}f(x)dx

With x_i^*\in[(i-1)/\sqrt{n}, i/\sqrt{n}].
I thought the interval is [0, ∞) since for each n, we have {0, 1/√n, 2/√n, ..., n/√n} a partition of [0, ∞), is that right?
Now I would need to find x_i* and f just by guessing?

 

[quasar987]

The theorem concerns linking Riemann sum and integral concerns definite integrals. [0,∞) is not a valid interval.

But

\frac{1}{\sqrt{n^2+in}}=\frac{1}{n}\frac{1}{\sqrt{1+i/n}}

 

[springo]

OK, now it all makes sense :)
\lim_{n\rightarrow\infty}\left(\frac{1}{\sqrt{n^2+ n}} + \frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+3n}}+...+\frac{1}{\sqrt{n^2+nn}} \right) = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{n}\frac{1}{\sqrt{1+\frac{i}{n}}} =\int_{0}^{1}\frac{dx}{\sqrt{1+x}}=\left[ 2\sqrt{1+x}\right]_0^1=2\sqrt{2}-2

 

[quasar987]

Well done.

 

[JG89]

I don't mean to hijack this thread, but I have a question for quasar.

In this step right here: \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{n} \frac{1}{\sqrt{1+\frac{i}{n}}} =\int_{0}^{1}\frac{dx}{\sqrt{1+x}}=\left[ 2\sqrt{1+x}\right]_0^1,

Is the integral from 0 to 1 because in the summation, when i = 1 then the term 1/n = 1 and as i tends to infinity, the term 1/n will tend to 0?

 

[quasar987]

The integral is from 0 to 1 because i/n is some point taken from the subinterval [(i-1)/n, i/n] of the partition {0, 1/n, 2/n, ..., n/n=1} of [0,1]. It's all explained in post #3.