- #1
isukatphysics69
- 453
- 8
Homework Statement
in title
Homework Equations
n = 2,3,4...
The Attempt at a Solution
n!/(n-2)! = n!/(n!(n-2)) = 1/(n-2) lim _{n->∞} = 1/∞ = 0 so sequence converges
Incorrect
Your first equality is wrong. Write out both the numerator and denominator of your original fraction.isukatphysics69 said:Homework Statement
in title
Homework Equations
n = 2,3,4...
The Attempt at a Solution
n!/(n-2)! = n!/(n!(n-2)) = 1/(n-2) lim _{n->∞} = 1/∞ = 0 so sequence converges
Incorrect
i don't understand, so the original is n!/((n-2)!)LCKurtz said:Your first equality is wrong. Write out both the numerator and denominator of your original fraction.
using n = 10 as an exampleLCKurtz said:Write out the factors of both numerator and denominator to see what cancels.
There is no factor of n! in the denominator.isukatphysics69 said:so i thought i can factor out an n! from the denominator?
As you wrote it, it looks like you factored 1/(n - 2) out of the limit expression. This isn't valid, because the limit is as n is changing. What you wrote is a little like writing 25 √, with nothing under the radical.isukatphysics69 said:1/(n-2) lim _{n->∞} = 1/∞
This is not a homework, i am studying for final exam. but general n i will attempt quickly so i would sayLCKurtz said:Yes, but now do it for general n, especially if this is a homework problem you are going to hand in.
Which makes it schoolwork.isukatphysics69 said:This is not a homework, i am studying for final exam.
No -- that's what you started with. After your simplification, how do you end up with the same thing you started with?isukatphysics69 said:(2*3*4*5*6...*n)/(1*2*3*4*5*6...(n-2))
cancel out everything until n leaving just
n!/(n-2)!
isukatphysics69 said:where n!>(n-2)! so diverges?
wait i mean it will leave justMark44 said:Which makes it schoolwork.
No -- that's what you started with. After your simplification, how do you end up with the same thing you started with?
No.isukatphysics69 said:wait i mean it will leave just
n/(n-2) after the factorials cancel
yes, so n(n - 1)(n - 2)! / (n-2)!Mark44 said:No.
Isn't n! = n(n - 1)(n - 2)! ?
Yes. Now wasn't that easier?isukatphysics69 said:yes, so n(n - 1)(n - 2)! / (n-2)!
so cancel the (n-2)! leaving just n(n-1)
so that would be inf * inf so the sequence divergesMark44 said:Yes. Now wasn't that easier?
So ##\lim_{n \to \infty}\frac{n!}{(n - 2)!} = \lim_{n \to \infty} n(n - 1) = ##?
And what does this say about your sequence?
Yesisukatphysics69 said:so that would be inf * inf so the sequence diverges
hey do you agree with what i am doing here to prove that the series n/(8n^3+6n^2-7) converges, i am using a limit comparison test n^2/(8n^3+6n^2-7)Mark44 said:Yes
No, I don't agree. When you use the limit comparison test, you are comparing the series you're working with against another series whose behavior you know.isukatphysics69 said:hey do you agree with what i am doing here to prove that the series n/(8n^3+6n^2-7) converges, i am using a limit comparison test n^2/(8n^3+6n^2-7)
i will now take the limit at infinity so i only care about the leading terms
n^{2}/8n^{3} so lim n approaching inf of n/(8n) = 0. this shows that a larger series converges since the infinite limit 0 so the smaller original series also converges
ok, ill be backMark44 said:No, I don't agree. When you use the limit comparison test, you are comparing the series you're working with against another series whose behavior you know.
The series you picked to compare against is a divergent series, being comparable to ##\sum \frac 1 n##, or more specifically, ##\sum \frac 1 {8n}##, which also diverges. I'm pretty sure you didn't know that.
You should rethink the series you're going to compare against.
wow, i think i can just use n/8n^3 as my comparison series. this would be 1/8n^2 which by the p series test i know converges since p>1Mark44 said:No, I don't agree. When you use the limit comparison test, you are comparing the series you're working with against another series whose behavior you know.
The series you picked to compare against is a divergent series, being comparable to ##\sum \frac 1 n##, or more specifically, ##\sum \frac 1 {8n}##, which also diverges. I'm pretty sure you didn't know that.
You should rethink the series you're going to compare against.
Much better choice.isukatphysics69 said:wow, i think i can just use n/8n^3 as my comparison series. this would be 1/8n^2 which by the p series test i know converges since p>1
Sure, that works.isukatphysics69 said:n/8n^3 > n/(8n^3+6n^2-7)
since the larger one converges, the smaller one must also converge
Since this is a new problem, please start a new thread.isukatphysics69 said:i have another one here that i think i solved proper but unsure.
from n=1 to inf
(cos(n*pi)) / (n^2)
|cos(n*pi)| <= 1
by comparison i can take the series 1/(n^2) which is > (cos(n*pi)) / (n^2)
by the p series test 1/(n^2) converges since p > 1
so by comparison (cos(n*pi)) / (n^2) must converge absolutely since |(cos(n*pi))| / (n^2) < 1/(n^2)
The formula for determining if a sequence converges or diverges is to take the limit as n approaches infinity of the sequence. If the limit exists and is a finite number, the sequence converges. If the limit does not exist or is infinity, the sequence diverges.
If the limit of the sequence as n approaches infinity is a finite number, the sequence is convergent. If the limit does not exist or is infinity, the sequence is divergent.
A convergent sequence has a finite limit as n approaches infinity, meaning the terms in the sequence get closer and closer to a specific number. A divergent sequence does not have a finite limit and the terms in the sequence either increase or decrease without bound.
To determine if the sequence n/(n-2) is convergent or divergent, you can plug in larger and larger values for n and see if the terms in the sequence approach a finite number or increase/decrease without bound. Alternatively, you can take the limit as n approaches infinity and see if the limit exists and is a finite number.
No, a sequence cannot be both convergent and divergent. A sequence can only have one limit as n approaches infinity, so it can either converge or diverge, but not both at the same time.