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prove the inequalities (n!)^2<_ n^n.n!<(2n)!

Asked by paritosh jain 8th August 2012, 10:57 AM
Answered by Expert
Answer:
(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.
Answered by Expert 8th August 2012, 2:38 PM
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