# Determine if sequence converges or diverges n/(n-2)

• isukatphysics69

in title

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

in title

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
Your first equality is wrong. Write out both the numerator and denominator of your original fraction.
i don't understand, so the original is n!/((n-2)!)
so i thought i can factor out an n! from the denominator?

Write out the factors of both numerator and denominator to see what cancels.

• isukatphysics69
Write out the factors of both numerator and denominator to see what cancels.
using n = 10 as an example

(10*9*8*7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1) leaving just 10*9 = means diverges at infinity

Yes, but now do it for general n, especially if this is a homework problem you are going to hand in.

so i thought i can factor out an n! from the denominator?
There is no factor of n! in the denominator.
Also, this isn't a good way to write things.
1/(n-2) lim n->∞ = 1/∞
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.

One last thing -- ∞ can't be used in arithmetic expressions, so 1/∞ is meaningless. However, you can say ##\lim_{n \to \infty} \frac 1 n = 0##.

• isukatphysics69
Yes, but now do it for general n, especially if this is a homework problem you are going to hand in.
This is not a homework, i am studying for final exam. but general n i will attempt quickly so i would say
(2*3*4*5*6...*n)/(1*2*3*4*5*6...(n-2))
cancel out everything until n leaving just
n!/(n-2)! where n!>(n-2)! so diverges?

This is not a homework, i am studying for final exam.
Which makes it schoolwork.

(2*3*4*5*6...*n)/(1*2*3*4*5*6...(n-2))
cancel out everything until n leaving just
n!/(n-2)!
No -- that's what you started with. After your simplification, how do you end up with the same thing you started with?
isukatphysics69 said:
where n!>(n-2)! so diverges?

• isukatphysics69
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?
wait i mean it will leave just
n/(n-2) after the factorials cancel

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wait i mean it will leave just
n/(n-2) after the factorials cancel
No.

Isn't n! = n(n - 1)(n - 2)! ?

• isukatphysics69
No.

Isn't n! = n(n - 1)(n - 2)! ?
yes, so n(n - 1)(n - 2)! / (n-2)!
so cancel the (n-2)! leaving just n(n-1)

yes, so n(n - 1)(n - 2)! / (n-2)!
so cancel the (n-2)! leaving just n(n-1)
Yes. Now wasn't that easier?

So ##\lim_{n \to \infty}\frac{n!}{(n - 2)!} = \lim_{n \to \infty} n(n - 1) = ##?

• isukatphysics69
Yes. Now wasn't that easier?

So ##\lim_{n \to \infty}\frac{n!}{(n - 2)!} = \lim_{n \to \infty} n(n - 1) = ##?
so that would be inf * inf so the sequence diverges

so that would be inf * inf so the sequence diverges
Yes

• isukatphysics69
Yes
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
n2/8n3 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

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
n2/8n3 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
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.

• isukatphysics69
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.
ok, ill be back

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>1

n/8n^3 > n/(8n^3+6n^2-7)

since the larger one converges, the smaller one must also converge

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
Much better choice.
isukatphysics69 said:
n/8n^3 > n/(8n^3+6n^2-7)
since the larger one converges, the smaller one must also converge
Sure, that works.
You could also use the limit comparison test, comparing your series against ##\frac 1 {8n^2}##, which you know converges, being a multiple of a convergent p-series..

Your reasoning was much better this time. These problems about convergence/divergence are basically true/false questions, where you would have a 50% chance of getting it right merely my guessing. To show you really understand, you have to provide solid reasoning.

• isukatphysics69
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)

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)