Question

Simplifying factorials, limits.

Answer 1

\[\lim_{n \rightarrow \infty}\left| \frac{ (2n)! }{ (2n+2)! } \right|\]

Answer 2

How do you get 0?

Answer 3

(2n+2)! = (2n+2)(2n+2)2n!

Answer 4

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

Answer 5

cancel 2n! with the numerator, you get \(lim_{n\rightarrow \infty}\dfrac{1}{4n^2+8n+2}\)

Answer 6

how does (2n+2)! = (2n+2) (2n+2) 2n!
?

Answer 7

oh, my bad, (2n+2)(2n+1) 2n!

Answer 8

Ah, I see. if it was (2n-2)! instead of (2n+2)!, it would be (2n-2)(2n-1)2n!
?

Answer 9

Just to check my factorial understanding.

Answer 10

Yes, you are correct
\(\dfrac{2n!}{(2n+2)(2n+1)2n!}=\dfrac{1}{(2n+2)(2n+1)}\)

Answer 11

Okay thanks :) I got it now.