Question
Wed August 08, 2012

# prove the inequalities (n!)^2<_ n>

Wed August 08, 2012
(2n)! = (1.2.3.... n) (n + 1) (n + 2) ... (2n - 1) (2n)
> (n!) . nn   [Since, (n + r) > n for r = 1, 2, 3, ... n]
Therefore, (n!) . nn < (2n)!                        (A)

Also, (n!)2 = (1.2.3.... n) n! <_ nn . (n!)       (B)
[Since, r is less than equal to n for each r = 1, 2, 3 , ... n]

Thus, from (A) and (B), we get
(n!)<_ nn . (n!) < (2n)!

Hence, proved.
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