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

[isukatphysics69]

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

 

[LCKurtz]

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]

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?

 

[LCKurtz]

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

 

[LCKurtz]

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

 

[Mark44]

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?

​

 

[Mark44]

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?

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

 

[Mark44]

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)

 

[Mark44]

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 $\lim_{n \to \infty}\frac{n!}{(n - 2)!} = \lim_{n \to \infty} n(n - 1) =$?
And what does this say about your sequence?

 

[isukatphysics69]

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?

so that would be inf * inf so the sequence diverges

 

[Mark44]

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

 

[Mark44]

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

 

[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.

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

 

[Mark44]

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.

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)

 

[Mark44]

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)

Since this is a new problem, please start a new thread.

Thread closed.