Question

Prove for n e N, (n!)^2<_n^n.n! < (2n)!.​

Answer 1

Answer:

consider

(n!)

2

≤n

n

(n!)part

n≤n

(n−1)≤n

(n−2)≤n

.....

.....

.....

1≤n

∴n(n−1)(n−2)......1≤n.n.n......ntimes

n!≤n

n

multiply both sides by n!, we get:

(n!)

2

≤n

n

(n!)

Now consider the second part n

n

(n!)≤(2n!)

as

n≤2n

n≤(2n−1)

n≤(2n−2)

.....

.....

.....

≤[2n−(n−1)]

∴n.n.....ntimes≤2n(2n−1)(2n−2)........[2n−(n−1)]

n

n

≤2n(2n−1)(2n−2)........[n+1]

multiply both sides by n!, we get

n

n

n!≤2n(2n−1)(2n−2)........[n+1](n!)

n

n

n!≤(2n!)

Hence Proved!

Answer 2

Answer:

answer for the given problem is given