{ (2n-1)! / (2n+1)! }?

4 ⤊

{ (2n-1)! / (2n+1)! }?

How do u figure how the limit of this when it this sequence converges

2007-11-27 15:13:44 · 3 answers · asked by pnetecos 1 in Science & Mathematics ➔ Mathematics

sry i meant..how do u figure the limit of this when this sequence converges

2007-11-27 15:18:16 · update #1

3 answers

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

2007-11-27 21:42:56 · answer #1 · answered by Como 7 · 1⤊ 1⤋

(2n+1)! = (2n-1)! * 2n * (2n+1)
So
(2n-1)! / (2n+1)! = 1/(2n(2n+1))
So obviously the sequence converges to zero.
You should use a proper epsilon/delta proof for the convergence.

2007-11-27 15:18:58 · answer #2 · answered by Ian 6 · 1⤊ 1⤋

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

2007-11-27 15:19:45 · answer #3 · answered by sv 7 · 2⤊ 1⤋