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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.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)!)Your first equality is wrong. Write out both the numerator and denominator of your original fraction.
using n = 10 as an exampleWrite out the factors of both numerator and denominator to see what cancels.
There is no factor of n! in the denominator.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.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 sayYes, but now do it for general n, especially if this is a homework problem you are going to hand in.
Which makes it schoolwork.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?(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 justWhich 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.wait i mean it will leave just
n/(n-2) after the factorials cancel
yes, so n(n - 1)(n - 2)! / (n-2)!No.
Isn't n! = n(n - 1)(n - 2)! ?
Yes. Now wasn't that easier?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 divergesYes. 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?
Yesso 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)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.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 backNo, 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>1No, 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.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.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)