Follow Math Help Forum on Facebook and Google+
Mistaken post. I can't remove it.
Last edited by Sudharaka; February 2nd 2010 at 04:44 PM.
?
(2)(4)(6)...(2n-2)(2n)= (2*1)(2*2)(2*3)...(2*(n-1))(2*n)= . 1(3)(5)...(2n-1)= so That in the denominator "dominates" is enough to show that the limit is 0.
Originally Posted by HallsofIvy (2)(4)(6)...(2n-2)(2n)= (2*1)(2*2)(2*3)...(2*(n-1))(2*n)= . 1(3)(5)...(2n-1)= so That in the denominator "dominates" is enough to show that the limit is 0. Dear HallsofIvy, I think you have made a slight error. The last line must be,
View Tag Cloud