Limits and Expression

[car0le_la]

If n is a positive integral then lim n->infinity 1/n[(1/n)²+(2/n)²+ ... + (n-1/n)²] can be expressed as...
a) ∫ 0 to 1 (1/x²) dx
b) ∫ 0 to 1 (x²) dx
c) ∫ 0 to 1 (2/x²) dx
d) ∫ 0 to 2 (x²) dx
Can someone please explain the overall thought process and what I would have to do if I came across another problem like this but with a different equation? I understand the concept of how integrals work but don't understand how it connects to the use of limits.

 

[Cubist]

If n is a positive integral then...

Hi, welcome to FMH !

Are you absolutely sure that the question uses the word "integral", could it perhaps be "integer"? The latter would make much more sense! It could be a typo in the question. Anyway, I recommend that you continue the question making the assumption that it says integer!

Just let me double check that the limit looks like this...

[math] \lim_{n \to \infty} \frac{1}{n} \left(\left(\frac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2 + \ldots + \left(\frac{n-1}{n}\right)^2\right) [/math]
I computed the above for n=100, then 1000, and it seems to be converging to a certain fraction which is also the result of one of the definite integrals.

 

[Dr.Peterson]

If n is a positive integer then lim n->infinity 1/n[(1/n)²+(2/n)²+ ... + (n-1/n)²] can be expressed as...
a) ∫ 0 to 1 (1/x²) dx
b) ∫ 0 to 1 (x²) dx
c) ∫ 0 to 1 (2/x²) dx
d) ∫ 0 to 2 (x²) dx
Can someone please explain the overall thought process and what I would have to do if I came across another problem like this but with a different equation? I understand the concept of how integrals work but don't understand how it connects to the use of limits.

Here's how I'd start:

We are given [MATH]\lim_{n \to \infty} \frac{1}{n} \left(\left(\frac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2 + \ldots + \left(\frac{n-1}{n}\right)^2\right)[/MATH]
We compare this to the definition of the definite integral (and in particular the Riemann sum), and see that the latter involves a summation of terms, each of which is something times [MATH]\Delta x[/MATH], which is getting very small in the limit. Since each term in the given expression is multiplied by [MATH]\frac{1}{n}[/MATH], maybe that corresponds to [MATH]\Delta x[/MATH].

Now, if that's [MATH]\Delta x[/MATH], what corresponds to [MATH]x[/MATH] itself? The first value of [MATH]x[/MATH] is the lower limit, and then each value in the sum is made by adding [MATH]\Delta x[/MATH] to the previous one. It looks like the [MATH]\frac{k}{n}[/MATH] within each term of our sum (with [MATH]k[/MATH] varying from [MATH]1[/MATH] to [MATH]n-1[/MATH]) is a value of [MATH]x[/MATH].

Can you now write a Riemann sum that corresponds to the sum we are given? Which integral will it correspond to?

 

[Steven G]

It must be integer and not integral.

I agree with Cubist and suggest that you think integral even if you did not have choices to pick from.

 

[car0le_la]

Here's how I'd start:

We are given [MATH]\lim_{n \to \infty} \frac{1}{n} \left(\left(\frac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2 + \ldots + \left(\frac{n-1}{n}\right)^2\right)[/MATH]
We compare this to the definition of the definite integral (and in particular the Riemann sum), and see that the latter involves a summation of terms, each of which is something times [MATH]\Delta x[/MATH], which is getting very small in the limit. Since each term in the given expression is multiplied by [MATH]\frac{1}{n}[/MATH], maybe that corresponds to [MATH]\Delta x[/MATH].

Now, if that's [MATH]\Delta x[/MATH], what corresponds to [MATH]x[/MATH] itself? The first value of [MATH]x[/MATH] is the lower limit, and then each value in the sum is made by adding [MATH]\Delta x[/MATH] to the previous one. It looks like the [MATH]\frac{k}{n}[/MATH] within each term of our sum (with [MATH]k[/MATH] varying from [MATH]1[/MATH] to [MATH]n-1[/MATH]) is a value of [MATH]x[/MATH].

Can you now write a Riemann sum that corresponds to the sum we are given? Which integral will it correspond to?

Would it be b? I see that every number within the Riemann sum seems to have the values squared. Following the idea that delta x is 1 that would mean that the integral is from 0 to 1. But this is more or so a rough concept. Also how would I double check my work to see if I got this right?

 

[car0le_la]

Hi, welcome to FMH !

Are you absolutely sure that the question uses the word "integral", could it perhaps be "integer"? The latter would make much more sense! It could be a typo in the question. Anyway, I recommend that you continue the question making the assumption that it says integer!

Just let me double check that the limit looks like this...

[math] \lim_{n \to \infty} \frac{1}{n} \left(\left(\frac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2 + \ldots + \left(\frac{n-1}{n}\right)^2\right) [/math]
I computed the above for n=100, then 1000, and it seems to be converging to a certain fraction which is also the result of one of the definite integrals.

This approach of just solving for the area under the curve of each definite integral given and then just plugging in a large value within the limit equation seems like a good technique.
so the area under the curve of each integral respectively
a- -----
b-1/3
c- -----
d-8/3
Would it work I was given another variation of this type of problem? What would I do if I wasn't given a calculator as solving for the limit with larger numbers would take some time.

 

[Cubist]

In post #2 I used a numerical method to quickly determine that the question made sense (that there is a likely candidate in the possible answers). This method should NOT be used to answer the question in a test/ exam since the question says "can be expressed as", not "is numerically close to". You are right that b (1/3) seems to be the correct answer. But to get the marks I think you need a stronger argument/ method...

So I'd try looking further into the method suggested by Dr.Peterson in post #3.

EDIT: A numerical method can sometimes be used as a double check - but beware that convergence can be slow sometimes, and doing n=1000 would not be feasible without a programmable device

 

[pka]

\(\lim_{n \to \infty}\displaystyle{ \dfrac{1}{n} \left(\left(\dfrac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2 + \ldots + \left(\dfrac{n-1}{n}\right)^2\right)}\)

I want to comment on this by way of what some authors call special sums.
\(\displaystyle{ \dfrac{1}{n} \left(\left(\dfrac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2 + \ldots + \left(\dfrac{n-1}{n}\right)^2\right)}\) becomes
\(\displaystyle{ \dfrac{1}{n^3} \left(1^2+2^2+\cdots+(n-1)^2\right)}=\dfrac{1}{n^3}\sum\limits_{k = 1}^{n - 1} {{k^2}} = \frac{{(n - 1)n(2n - 1)}}{6n^3}\)

 

[pka]

I want to comment on this by way of what some authors call special sums.
\(\displaystyle{ \dfrac{1}{n} \left(\left(\dfrac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2 + \ldots + \left(\dfrac{n-1}{n}\right)^2\right)}\) becomes
\(\displaystyle{ \dfrac{1}{n^3} \left(1^2+2^2+\cdots+(n-1)^2\right)}=\dfrac{1}{n^3}\sum\limits_{k = 1}^{n - 1} {{k^2}} = \frac{{(n - 1)n(2n - 1)}}{6n^3}\)

See HERE to verify results.
The end result gives \(\dfrac{2n^3-3n^2+n}{6n^3}\) if \(n\to\infty\) what is the limit?

 

[car0le_la]

See HERE to verify results.
The end result gives \(\dfrac{2n^3-3n^2+n}{6n^3}\) if \(n\to\infty\) what is the limit?

The limit would be 1/3 because if you attain look at the degree of both the top and bottom being the same you using the leading coefficient of the top and bottom, which is 2/6 which simplifies to 1/3

 

[car0le_la]

I want to comment on this by way of what some authors call special sums.
\(\displaystyle{ \dfrac{1}{n} \left(\left(\dfrac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2 + \ldots + \left(\dfrac{n-1}{n}\right)^2\right)}\) becomes
\(\displaystyle{ \dfrac{1}{n^3} \left(1^2+2^2+\cdots+(n-1)^2\right)}=\dfrac{1}{n^3}\sum\limits_{k = 1}^{n - 1} {{k^2}} = \frac{{(n - 1)n(2n - 1)}}{6n^3}\)

thanks!