# Square of n factorial is greater than n to the power n

Hi

I need some help figuring out how to do this problem:

Prove that for every natural number n, $$(n!)^{2} > n^n$$

By rewriting this as
$$(n!)^{\frac{1}{n}} > \sqrt{n}$$
I can see that I have to prove that given a sequence of natural numbers {1, 2, 3, ...., n}, the geometric mean of n numbers of this sequence is greater than the geometric mean of its extreme terms, i.e. 1 and n. I get stuck however, when I attempt a noninductive proof (even an inductive proof for that matter involves handling binomial expressions which are large but it is perhaps still possible). I don't see how to go on now that I have the rearranged inequality even though in its present form and significance, it looks pretty obvious.

I would be grateful if someone could offer a suggestion to help me analyze this.

Thanks and cheers
Vivek

Tide
Homework Helper
How about trying a proof by induction?

If you do that then I think it reduces to proving
$$\frac { \left(1 + \frac {1}{n}\right)^n}{n+1} \leq 1$$

Hello Tide

I've got to do it noninductively :-)

Cheers
Vivek

Gokul43201
Staff Emeritus
Gold Member
Just a thought - don't have time now to look at this closely - but what is the nature of the error in Stirling's approximation (log(n!) ~ nlogn-n) ? You could look into this...

jcsd
Gold Member
I'll take a guess too: If you take a look at the Stirling series you should be able to pull out a formula that's guraenteed to give an approximation greater than or less than the actual value.

Tide
Homework Helper
Stirling won't work if he has to prove it for all natural numbers.

It is not true for 1 or 2 anyway. 3 is the first integer it is true for. Sterling's formula says: n! approximately $$\frac{\sqrt(2n\pi)*n^n}{e^n}$$.This should help.

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Thank you all for your help. I would like to point out however, that this problem is from a book called Higher Algebra by Hall and Knight (chapter on Inequalities). As the stirling approximation has not been mentioned in the theory sections, I believe the authors expect students to solve it some other way.

Thanks and cheers
Vivek

mathwonk
Homework Helper
2020 Award
well just lookinbg at it, the lhs is being multiplied by n^2 each time you go from the case for n-1 to the case for n, and the rhs is being multiplied by an element of a sequence that converges to en. so the lhs is obviously increasing much faster.

Gokul43201
Staff Emeritus
Gold Member
Here's another quick suggestion. This one may go nowhere too :

Consider the numbers 1, 2, 3, ..., n. Use the AM > GM inequality on them. See if that goes anywhere(or even 1^2, 2^2, 3^2, ..., n^2). Looks to me like the inequalities might point the wrong way...uh oh !

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1) Hall and Knight seem to use > to mean greater than or equal to

2)
maverick280857 said:
I can see that I have to prove that given a sequence of natural numbers {1, 2, 3, ...., n}, the geometric mean of n numbers of this sequence is greater than the geometric mean of its extreme terms, i.e. 1 and n. I get stuck however, when I attempt a noninductive proof
So if you can prove n*1>=n, (n-1)*2>=n and so on then that will do it.

maverick280857 said:
I've got to do it noninductively :-)
That sucks... what's wrong with an inductive proof?

Gokul43201
Staff Emeritus
Gold Member
chronon said:
So if you can prove n*1>=n, (n-1)*2>=n and so on then that will do it.
Ha ha, simple as that !

Just prove : (n-k)(k+1) >= n for all k in [0,n-1].

Nice !

Now I feel like heading into the nearest wall. :

Hello everyone

Thanks for the suggestions. I think I have it figured out now :-)

Well, there's nothing wrong with an inductive proof but you cannot say that induction is the nicest way to do things. I mean there's got to be a "formal" proof or a better looking noninductive proof. If I could do things two ways, I'd prefer a noninductive proof because I need to put my mind more into it. Normally induction just involves using the proposition for k to prove it for (k+1).

Thanks anyway.

Cheers
Vivek

robert Ihnot said:
It is not true for 1 or 2 anyway. 3 is the first integer it is true for. Sterling's formula says: n! approximately $$\frac{\sqrt(2n\pi)*n^n}{e^n}$$.This should help.
As a side note there is a better approximation of n! :

[√ (2nπ)]*nn*e-n+(1/12n) (1)

Stirling formula can be deduced from a formula (attributed to Euler) where an infinite series of a function in x is approximated by its integral (integral from f(t)dt,between 0 and 'x')+a series of factors depending on the derivatives of the function+a residue.In our case after taking the logarithm from n! this function is f(x)=lnx.If I remember well only the first derivatives and the rest of the first order are considered,the other terms are ignored.The residue of order 1 is demonstrated to converge at a certain finite limit C.I do not remember it exactly but the fact is that finally results (1).Stirling's formula results directly from there by ignoring the factor (1/12n).

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