how do you get

(2n-1)! / (2n+1)! to equal to

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

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- Apr 26th 2007, 10:51 AMjephfactorials
how do you get

(2n-1)! / (2n+1)! to equal to

(2n-1)! / (2n+1)(2n)(2n-1)! - Apr 26th 2007, 11:09 AMThePerfectHacker
- Apr 26th 2007, 11:43 AMjeph
well i thought it had something to do with calculus...its the first time I've seen the ! sign people used in a problem

the question was:

determine whether the sequence converges or diverges. If it converges, find the limit.

i just needed to know how that step was done... - Apr 26th 2007, 11:54 AMJhevon
- Apr 26th 2007, 12:16 PMJhevon
we're talking about a SEQUENCE and not a SERIES right. and i suppose you want the limit as n-->oo

well in that case:

yes, the sequence does converge--to zero

lim{n-->oo}[(2n-1)!/(2n+1)! ] = lim{n-->oo}[(2n-1)!/(2n+1)(2n)(2n - 1)!]

.........................................= lim{n-->oo}[1/(2n + 1)(2n)]

.........................................= 0