Convergence of sequence ++

[SkandinavianNeedsHelp]

Hi! Need some help. Sorry for bad english, hope I can make myself understandable.

Problem:







Using L'Hopitals rule i find the the limit in the first problem is 0.

I think i'm able to use the mean value theorem to show that an/(n-1)=(e^-nc)*c. f'(c)=(f(b)-f(a))/(b-a) where f(b) - f(a) = an because f(a) is 0 (integral from 1 to 1) (??). Am I thinking correct here?

I have no idea of how to approach the rest of the exersice. I can see that the left side in the inequality equals an and the right side the sequence in the first problem , but i have no idea of how to proceed.

Thanks in advance.

 

[Steven G]

Yes, use the MVT for that part of the question (actually that was given)
For the inequality surely you can do something. Maybe divide by n-1 etc. Please show us your work.
For the limit question, maybe compute aⁿ and then take the limit?

 

[SkandinavianNeedsHelp]

Yes, use the MVT for that part of the question (actually that was given)
For the inequality surely you can do something. Maybe divide by n-1 etc. Please show us your work.
For the limit question, maybe compute aⁿ and then take the limit?

Thanks for your reply. I can upload my work later today when I get home. About the inequality I dont know the role of c, and I dont know what i want to achieve by "doing something". I'm pretty lost here

 

[SkandinavianNeedsHelp]

Find the limits:

[MATH]lim_{n \to \infty} \frac{n^4 -n^3}{e^n}[/MATH]

Using L'Hospital's rule I got that the limit equals to 0.

(1)
The sequence [MATH]a_n[/MATH] given by:
[MATH]a_n = \int_{1}^{t} e^{-nx} x^3 dx, \ n=1,2,\ldots [/MATH]
then [MATH]a_n[/MATH] is the function value [MATH]F(n)[/MATH], as:

[MATH]F(t) = \int_{1}^{t} e^{-nx} x^3 dx[/MATH]
(2)
Use the mean value theorem on the function [MATH]F(t)[/MATH] in the interval [MATH][1,n][/MATH] to show that: [MATH]\frac{a_n}{n-1} = e^{-nc} c^3, \ c\in (1, n)[/MATH]

I think I'm able to show this. Is it correct that [MATH]f'(c)=\frac{f(b)-f(a)}{b-a} = \frac{f(b)}{b-a}[/MATH] because [MATH]f(a) = o[/MATH] (integral from 1 to 1)?

(3)
Why is [MATH]e^{-nc} c^3 (n-1) \le e^{-n} n^{3} (n-1), c\in (1,n) [/MATH]
I understand this now since [MATH]c \lt n [/MATH]and [MATH](n-1)[/MATH] is always a positive number.


(4)
Use the inequality to show that the sequence [MATH]a_n[/MATH] converges on find the limit [MATH]lim_{a_n \to \infty}[/MATH]

Since the right side of the inequality converges to 0 (task 1), [MATH]lim_{a_n \to \infty} = e^{-nc} c^3 (n-1)[/MATH] = left side of inequality must converge to 0 (task 4) as well since it is a positive smaller number?

 

[Steven G]

Maybe I should have been more clear, sorry. When you said MVT, I thought you mean the mean value theorem for integrals.
The Mean Value Theorem for Integrals guarantees that for every definite integral, a rectangle with the same area and width exists.
That is the definite integral = (b-a)f(c) where the integral goes from a to b, c is in (a,b) and f(x) is the integrand.

 

[Steven G]

I think I'm able to show this. Is it correct that [MATH]f'(c)=\frac{f(b)-f(a)}{b-a} = \frac{f(b)}{b-a}[/MATH] because [MATH]f(a) = o[/MATH] (integral from 1 to 1)?

If the integral is from 1 to 1 would not that imply that a=b=1? Then you get f'(c)= 0/0???Also, why f(a)=0??

 

[SkandinavianNeedsHelp]

Maybe I should have been more clear, sorry. When you said MVT, I thought you mean the mean value theorem for integrals.
The Mean Value Theorem for Integrals guarantees that for every definite integral, a rectangle with the same area and width exists.
That is the definite integral = (b-a)f(c) where the integral goes from a to b, c is in (a,b) and f(x) is the integrand.

If the integral is from 1 to 1 would not that imply that a=b=1? Then you get f'(c)= 0/0???Also, why f(a)=0??

Theory described here: http://people.math.sc.edu/meade/Bb-CalcI-WMI/Unit5/HTML-GIF/MVTIntegral.html

MVT for integrals is just a rephrase of "regular" MVT.

I could have been more clear using capital letters. F(a) is an integral from a to a, which is 0. F(b) is an integral from a to b. So F(b)-F(a) = F(b). a =1 and b = n.

 

[Steven G]

Theory described here: http://people.math.sc.edu/meade/Bb-CalcI-WMI/Unit5/HTML-GIF/MVTIntegral.html

MVT for integrals is just a rephrase of "regular" MVT.

No, sorry but I disagree with your statement. Even the link you supplied disagreed with your statement. Your link says that The Mean Value Theorem for Integrals is a direct consequence of the Mean Value Theorem (for Derivatives) and the First Fundamental Theorem of Calculus.
Are you clear on what a consequence means? The MVT for Integrals being a consequence for the MVT for derivatives says NOTHING at all about the two having similar definitions

Here is a great example to consider. We know that the union of the Rational numbers and Irrational numbers make up the Real numbers. Now we have two theorems. One theorem says that the Reals have an uncountable number of numbers. A 2nd theorem theorem says that the Rationals are a countable set. (note that the union of two countable sets is still countable). Now a consequence of these two theorems is that the irrationals must be uncountable.

Another example: There is a theorem that states that the even integers are countable. A consequence of that is that the odd integers are also countable (since there exists a bijection between the evens and the odds).

A consequence of my not going to school tomorrow is that I can go to the movies. Is skipping school and going to the movies really the same?

The MVT for integrals and the MVT for derivative are not the same. Just think about what they each say. Does the definition of one really imply the other one exactly??

 

[SkandinavianNeedsHelp]

No, sorry but I disagree with your statement. Even the link you supplied disagreed with your statement. Your link says that The Mean Value Theorem for Integrals is a direct consequence of the Mean Value Theorem (for Derivatives) and the First Fundamental Theorem of Calculus.
Are you clear on what a consequence means? The MVT for Integrals being a consequence for the MVT for derivatives says NOTHING at all about the two having similar definitions

Here is a great example to consider. We know that the union of the Rational numbers and Irrational numbers make up the Real numbers. Now we have two theorems. One theorem says that the Reals have an uncountable number of numbers. A 2nd theorem theorem says that the Rationals are a countable set. (note that the union of two countable sets is still countable). Now a consequence of these two theorems is that the irrationals must be uncountable.

Another example: There is a theorem that states that the even integers are countable. A consequence of that is that the odd integers are also countable (since there exists a bijection from the evens to the odds).

A consequence of my not going to school tomorrow is that I can go to the movies. Is skipping school and going to the movies really the same?

The MVT for integrals and the MVT for derivative are not the same. Just think about what they each say. Does the definition of one really imply the other one exactly??

Okey, so I was worng. It is a direct consequence of the MVT.

But do you still disagree with my calculations?

 

[Steven G]

You say f(a)=0. Where is anything said about a function named f? I looked twice at your initial post and see nothing about f. If you choose you can introduce a new function, like f, but you really need to define it for us.

 

[SkandinavianNeedsHelp]

You say f(a)=0. Where is anything said about a function named f? I looked twice at your initial post and see nothing about f. If you choose you can introduce a new function, like f, but you really need to define it for us.

f(a)=F(a)=F(1)