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

  • #1
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Hi

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

Prove that for every natural number n, [tex](n!)^{2} > n^n[/tex]

By rewriting this as
[tex](n!)^{\frac{1}{n}} > \sqrt{n}[/tex]
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
 

Answers and Replies

  • #2
Tide
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How about trying a proof by induction?

If you do that then I think it reduces to proving
[tex]\frac { \left(1 + \frac {1}{n}\right)^n}{n+1} \leq 1[/tex]
 
  • #3
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Hello Tide

I've got to do it noninductively :-)

Cheers
Vivek
 
  • #4
Gokul43201
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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...
 
  • #5
jcsd
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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.
 
  • #6
Tide
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Stirling won't work if he has to prove it for all natural numbers.
 
  • #7
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It is not true for 1 or 2 anyway. 3 is the first integer it is true for. Sterling's formula says: n! approximately [tex] \frac{\sqrt(2n\pi)*n^n}{e^n}[/tex].This should help.
 
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  • #8
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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
 
  • #9
mathwonk
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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.
 
  • #10
Gokul43201
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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 ! :confused:
 
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  • #11
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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.
 
  • #12
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maverick280857 said:
I've got to do it noninductively :-)
That sucks... what's wrong with an inductive proof?
 
  • #13
Gokul43201
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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 ! :smile:

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

Nice ! :smile:

Now I feel like heading into the nearest wall. : :cry:
 
  • #14
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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
 
  • #15
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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 [tex] \frac{\sqrt(2n\pi)*n^n}{e^n}[/tex].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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