Thread: factorials

how do you get

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

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

Originally Posted by jeph

how do you get

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

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

WHAT DOES THIS HAVE TO DO WITH CALCULUS!!!

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

= (2n+1)(2n) * [ (2n-1)*(2n-2)*...*(2)(1) ]

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

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...

Originally Posted by jeph

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...

TPH is right. just remember that the factorial of an integer is the product of all the integers from 1 up to that integer

Originally Posted by jeph

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...

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