Find a limit with a Riemann sum

[bluemath]

Hello,

I must find this limit with a Riemann sum. I see nothing, then is it possible please to have some help for starting please ?


lim x---> +inf x² (1/(x+1)³ + 1/(x+2)³ + ....+ 1/(2x)³ )

My reasoning :

When x is big, x² (1/(x+1)³) = ~ x²(1/x³) = 1/x

Then the sumn, we have x times ~[ (1/x + 1/8x)/2] when x is big.

Thanks

 

[pka]

I must find this limit with a Riemann sum. I see nothing, then is it possible please to have some help for starting please ?
lim x---> +inf x² (1/(x+1)³ + 1/(x+2)³ + ....+ 1/(2x)³ )
My reasoning :
When x is big, x² (1/(x+1)³) = ~ x²(1/x³) = 1/x
Then the sumn, we have x times ~[ (1/x + 1/8x)/2] when x is big.

@bluemath, this is sadly an example of how NOT to ask a question here.
What is the actual original? I have no idea how the limit you posted relates to any Riemann Sum.

In any case, \(\displaystyle \large\displaystyle {{\lim _{N \to \infty }}\sum\limits_{k = 1}^N {\left( {\frac{{{N^2}}}{{{{(N + k)}^3}}}} \right)} \ge {\lim _{N \to \infty }}\sum\limits_{k = 1}^N {\frac{1}{{8N}}}} \)

 

[bluemath]

@bluemath, this is sadly an example of how NOT to ask a question here.
What is the actual original? I have no idea how the limit you posted relates to any Riemann Sum.

I really don't understand the problem.

My question is to find the limit and we had an help who say : " this is a hidden Rieman sum "

So what is your problem exactly ? What is so sad ?

By the way, thanks for your answer.

 

[pka]

I really don't understand the problem.
My question is to find the limit and we had an help who say : " this is a hidden Riemann sum "
So what is your problem exactly ? What is so sad ?

I don't have a problem. You do. I understand a great deal about Riemann integration, or what is called refinement-partition integrals. It is impossible to explain an approximating sum without knowing the intergrand.

What is sad? The fact someone who seems to know noting about it (there is no hidden sum) would mislead you in the pretense of helping.

 

[Ishuda]

Hello,

I must find this limit with a Riemann sum. I see nothing, then is it possible please to have some help for starting please ?


lim x---> +inf x² (1/(x+1)³ + 1/(x+2)³ + ....+ 1/(2x)³ )

My reasoning :

When x is big, x² (1/(x+1)³) = ~ x²(1/x³) = 1/x

Then the sumn, we have x times ~[ (1/x + 1/8x)/2] when x is big.

Thanks

Suppose we had the function
f(t) = \(\displaystyle \frac{1}{t^3}\)
and we want to integrate this from 1 to some T. Well choose the number of intervals N, let
\(\displaystyle \Delta\, =\, \frac{T-1}{N}\)
and we have a Riemann sum of
\(\displaystyle \int_1^T\, f(t) dt\) ~ R(T;N) = \(\displaystyle \Sigma_{k=0}^{k=N}\, f(t_k)\Delta\)
Since
\(\displaystyle t_k\, = 1\, +\, k \Delta\)
we have
R(T;N) = \(\displaystyle \Sigma_{k=0}^{k=N}\, \frac{1}{[1\, +\, k \Delta]^3} \Delta\)
Substitute for \(\displaystyle \Delta\) and determine how this is related to your problem.

 

[Ishuda]

I don't have a problem. You do. I understand a great deal about Riemann integration, or what is called refinement-partition integrals. It is impossible to explain an approximating sum without knowing the intergrand.

What is sad? The fact someone who seems to know noting about it (there is no hidden sum) would mislead you in the pretense of helping.

You are being so considerate again. I thank you for your thoughts.

 

[bluemath]

What is sad? The fact someone who seems to know noting about it (there is no hidden sum) would mislead you in the pretense of helping.

Probably a translation problem for " hidden ". I don't know if " disguised " is better.

 

[bluemath]

Suppose we had the function
f(t) = \(\displaystyle \frac{1}{t^3}\)
and we want to integrate this from 1 to some T. Well choose the number of intervals N, let
\(\displaystyle \Delta\, =\, \frac{T-1}{N}\)
and we have a Riemann sum of
\(\displaystyle \int_1^T\, f(t) dt\) ~ R(T;N) = \(\displaystyle \Sigma_{k=0}^{k=N}\, f(t_k)\Delta\)
Since
\(\displaystyle t_k\, = 1\, +\, k \Delta\)
we have
R(T;N) = \(\displaystyle \Sigma_{k=0}^{k=N}\, \frac{1}{[1\, +\, k \Delta]^3} \Delta\)
Substitute for \(\displaystyle \Delta\) and determine how this is related to your problem.

Many thanks

 

[bluemath]

I thought of having understood. I tried but finally not succeeded.
And I didn't find how to add the opposite of cubes.
Exercise too much difficult for me.

 

[Ishuda]

I thought of having understood. I tried but finally not succeeded.
And I didn't find how to add the opposite of cubes.
Exercise too much difficult for me.

O.K. Let's start with the Riemann sum for
\(\displaystyle \int_1^T\, f(t) dt\)[/tex]
where
f(t) = \(\displaystyle \frac{1}{t^3}\)
We choose the number of intervals N and let
\(\displaystyle \Delta\, =\, \frac{T-1}{N}\)
and we can write the Riemann sum as
R(T;N) = \(\displaystyle \Sigma_{k=0}^{k=N}\, \frac{1}{[1\, +\, k \Delta]^3} \Delta\)
Now substitute for \(\displaystyle \Delta\) to get
R(T;N) = \(\displaystyle \Sigma_{k=0}^{k=N}\, \frac{1}{[1\, +\, k \frac{T-1}{N}]^3} \frac{T-1}{N}\)
= \(\displaystyle \Sigma_{k=0}^{k=N}\, \frac{N^3}{[N\, +\, k (T-1)]^3} \frac{T-1}{N}\)
= \(\displaystyle \Sigma_{k=0}^{k=N}\, \frac{(T-1)\, N^2}{[N\, +\, k (T-1)]^3}\)
= (T-1) N² \(\displaystyle \left[\frac{1}{[N]^3}\, + \frac{1}{[N\, + (T-1)]^3}\, + \frac{1}{[N + 2 (T-1)]^3}\, \frac{1}{[N + 3 (T-1)]^3}\, +\, ...\, \frac{1}{[N + N (T-1)]^3}\right]\)
= (T-1) N² \(\displaystyle \left[\frac{1}{[N]^3} + \frac{1}{[N + (T-1)]^3}\, + \frac{1}{[N + 2 (T-1)]^3}\, \frac{1}{[N + 3 (T-1)]^3} +\, ...\, + \frac{1}{[N\, T]^3}\right]\)
= (T-1) N² \(\displaystyle \frac{T-1}{N}\) + \(\displaystyle \left[\frac{1}{[N + (T-1)]^3}\, + \frac{1}{[N + 2 (T-1)]^3}\, \frac{1}{[N + 3 (T-1)]^3} +\, ...\, + \frac{1}{[N\, T]^3}\right]\)
Now as N goes to infinity, we can drop that first term since it will go to zero, let a=T-1 and we could write
\(\displaystyle \int_1^T\, f(t) dt\)
= \(\displaystyle \lim_{N \to \infty} a N^2\left[\frac{1}{[N + a]^3}\, + \frac{1}{[N + 2 a]^3}\, \frac{1}{[N + 3a]^3} +\, ...\, + \frac{1}{[(a+1)N]^3}\right]\)
What does a have to be to almost match your expression of
\(\displaystyle \lim_{x \to \infty}x^2\left[\frac{1}{[x + 1]^3}\, + \frac{1}{[x + 2]^3}\, \frac{1}{[x + 3]^3} +\, ...\, + \frac{1}{[2\, x]^3}\right]\)
and what do you have to do [multiply by something, add something, ...] to R(T;N) to make it the same?

 

[pka]

I must find this limit with a Riemann sum. I see nothing, then is it possible please to have some help for starting please ?
lim x---> +inf x² (1/(x+1)³ + 1/(x+2)³ + ....+ 1/(2x)³ )

After ten postings in this thread, I still do not know what the question really is.
I suspect the question is really \(\displaystyle \sum\limits_{k = 1}^\infty {\frac{{{N^2}}}{{{{\left( {N + k} \right)}^3}}}}=~? \)
And the expected answer is \(\displaystyle \int_1^2 {{t^{ - 3}}dt} \).

Until about ten years ago, that was a usual question on the finial test for AP calculus. (today calculators and CAS have made it obsolete). It was expected that the student should recognize that sum is an approximating sum for that integral.

We see that \(\displaystyle \displaystyle \int_1^2 {{t^{ - 3}}dt} \approx \sum\limits_{k = 1}^N {\frac{1}{{{{\left( {1 + \frac{k}{N}} \right)}^3}}}\frac{1}{N}} \). As \(\displaystyle N \to \infty \) that is a righthand Riemann Sum of \(\displaystyle f(t)=t^{-3}\) on the interval \(\displaystyle [1,2]\).

 

[bluemath]

Thanks a lot for you two last posts who have helping me a lot !

I'm now in the good way I think.

And sorry if all was not so clear but you answered to my questions :wink: