how can be proven that (n!)^2>n^n?

Question

n in N
Maybe by usin induction or by arythmetic ang geometric progresions!

for n>1 n in N

this may help:(1+2+...+n)/n>=(n!)^1/n
(1+2+...+n)/n=(n+1)/2>=n^1/2

Answer 1

♠ so that is (n!)^2>n^n or (n!)^2/n^n>1 ?????
♣ n!= n(n-1)(n-2)(n-3)( *** )*2*1;
n!*n! =(n·1)* ((n-1)·2)* ((n-2)·3) ***(1·n);
♦ (n!)^2/n^n = 1*(2-2/n)*(3-6/n)*(4-12/n) (***) (4-12/n)*(3-6/n)*(2-2/n)*1;
♥ since each factor in ♦ >=1, then ♦ >1;
■ e.g. n=4, then ♦ =1*(2-2/4)(3-6/4)(4-12/4)= 1*1.5*1.5*1 > 1;

Answer 2

What Mr. Razor given is a simple and correct proof but need some clarification.

(n!)^2 = [n * (n-1) * (n-2) * ..... 2 * 1] * [n * (n-1) * (n-2) * ..... 2 * 1]

The terms inside the [ ] is nothing but the expanded form of n!. There are n terms inside this. When the terms inside the second [ ] are taken in reverse order we get.

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

Re-write this by writing terms from each sets alternatively

n * 1 * (n-1) * 2 * (n-2) * 3 .... (n-x) * (x+1) .... 2 * (n-1) * 1 * n

Note that all these terms are in the form (n-x) * (x+1) where n > x.

This terms will be equal to n when x = 1 (or x = n-1) and in all other cases will be more than n.

So
n * 1 * (n-1) * 2 * (n-2) * 3 .... (n-x) * (x+1) .... 2 * (n-1) * 1 * n > n * n * ......n * n (n multiplied times).

Thus the proof.

Answer 3

Your assertation is only true for n>2. To prove it, start at n=3 which is true since 6^2 > 3^3.

Now the induction, given (n!)^2 > n^n, show that it holds for n+1.

Expand the two sides:
((n+1)!)^2 = (n+1)^2 * (n!)^2
(n+1)^(n+1) = (n+1) * (n+1)^n

So it needs to be proven that:
(n+1)^2 * (n!)^2 > (n+1) * (n+1)^n

Expand (n+1)^n by the binomial expansion:

Numbering the terms from 0 to n, the kth term is:

n! /(k! * (n-k)!) * n^(n-k)

Look at the term n!/(n-k)! this is n * (n-1) * (n-2) ...... (n-k+1)
This is the product of k terms each no greater than n so it follows that n!/(n-k)! < n^k. So each term in the expansion is less than n^k * n^(n-k) = n^n. There are n+1 terms so

(n+1)^n < (n+1)*n^n.

The item to be proven was reduced to:
(n+1)^2 * (n!)^2 > (n+1) * (n+1)^n

Combining this with the recent result of (n+1)^n < (n+1)*n^n.we can replace term on the right of (n+1)^n with the larger quantity of (n+1)*n^n

This results in:

(n+1)^2 * (n!)^2 > (n+1) * (n+1) * n^n

Dividing out (n+1)^2 gives: (n!)^2 > n^n

Which was true as the basis of the induction

Answer 4

I would try induction.

(n!)^2 > n^n

1) Base case:
Let n =3. Then (3!)^2 = (6)^2 = 36, and
3^3 = 27. So the formula holds true for n = 3.

2) Assume the formula holds true for up to n = k. That is,

(k!)^2 > k^k.

{We want to prove that [(k + 1)!]^2 > (k + 1)^(k + 1)}

But

[(k + 1)!]^2
= [(k + 1)k!]^2
= (k + 1)^2 (k!)^2

Therefore

(k + 1)^2 (k!)^2 > (k + 1)^2 (k^k)

(Still working on problem... in progress. Currently incomplete).

Answer 5

the answer is in the n!.

any number that has the ! after it makes it a whole different number

if n were to equal 6, then n! would = 6 * 5 * 4 * 3 * 2 * 1

Answer 6

You can prove (n!)^2>n^n like this.

Replace n with x. Now you get (x!)^2>x^x
It is a proven fact that (x!)^2 is always >x^x
Hence (n!)^2>n^n
Proved.

Answer 7

(n!)² = [ n*(n-1)*(n-2) ...]²

and

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

n^n = n * n * n * n ...

QED