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How do you do (n+2)(n-2)! ?

Do you just expand it? How do you make the ! disappear?
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by Airess3
Do you just expand it? How do you make the ! disappear?


I do not quite understand what you are trying to do ...
disappear?

(n+2)(n-2)! = (n+2)!/[(n+2)(n+1)n(n-1)]
Reply 2
Avatar for davros
davros
16
Original post by Airess3
Do you just expand it? How do you make the ! disappear?


I think you need to give us the full question so we can see what exactly you're trying to do!
Original post by TeeEm
I do not quite understand what you are trying to do ...
disappear?

(n+2)(n-2)! = (n+2)!/[(n+2)(n+1)n(n-1)]




I'm trying to find if this equation converges or diverges and so far, this is my working. I don't know if I need to expand it or do any more working or if I just substitute n tends to infinity into the equation will be enough working now.

In the answer booklet it said the final step should be lim n tends to infinity (n+2)^-1=0 and R(eal numbers) =infinity. x is such that R.

20150411_184540.jpg
(edited 9 years ago)
Reply 4
Avatar for davros
davros
16
Original post by Airess3
I'm trying to find if this equation converges or diverges and so far, this is my working. I don't know if I need to expand it or do any more working or if I just substitute n tends to infinity into the equation will be enough working now.

In the answer booklet it said the final step should be lim n tends to infinity (n+2)^-1=0 and R(eal numbers) =infinity. x is such that R.

20150411_184540.jpg


You seem to have made life unnecessarily difficult for yourself by trying to expand too far!

All you need to note is that (n+2)! = (n+2)(n+1)! and then you can cancel the (n+1)! straight away :smile:
Original post by davros
You seem to have made life unnecessarily difficult for yourself by trying to expand too far!

All you need to note is that (n+2)! = (n+2)(n+1)! and then you can cancel the (n+1)! straight away :smile:


I'm stuck, why does (n+2)! = (n+2)(n+1)! but not (n+2)(n)(n-2)! ?
Reply 6
Avatar for davros
davros
16
Original post by Airess3
I'm stuck, why does (n+2)! = (n+2)(n+1)! but not (n+2)(n)(n-2)! ?


I don't know where you've got the latter from because it's completely wrong!

n factorial means n times all the integers below it down to 1.

So n! = n x (n-1) x (n-2) x ... x 1 = n x (n-1)!

and (n+2)! = (n+2) x (n+1) x n x (n-1) x ... x 1 = (n+2) x (n+1)!
Original post by davros
I don't know where you've got the latter from because it's completely wrong!

n factorial means n times all the integers below it down to 1.

So n! = n x (n-1) x (n-2) x ... x 1 = n x (n-1)!

and (n+2)! = (n+2) x (n+1) x n x (n-1) x ... x 1 = (n+2) x (n+1)!


Ok, thanks!
Original post by davros
I don't know where you've got the latter from because it's completely wrong!

n factorial means n times all the integers below it down to 1.

So n! = n x (n-1) x (n-2) x ... x 1 = n x (n-1)!

and (n+2)! = (n+2) x (n+1) x n x (n-1) x ... x 1 = (n+2) x (n+1)!


ok, just to check, it's absolutely convergent?
Original post by Airess3
ok, just to check, it's absolutely convergent?


Yes the series is convergent.

It is absolutely convergent as well, but you do know they aren't the same thing? Absolute convergence is when ∑n=1∞∣an∣\sum_{n=1}^{\infty }|a_n| converges. It doesn't make a difference in this case because all of the terms are positive, but just make sure you don't confuse convergence and absolute convergence. :smile:

(Have a look in your notes for something like 'conditionally convergent' - when a series is convergent but not absolutely convergent. A good example of this is ∑n=1∞(−1)n−1n\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{n})
(edited 9 years ago)

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