Stirling's Formula: How does n relate to n^n.

[Ryuky]

Hello all.
I am doing a school paper on Stirling's formula and I want first to show how the factor e comes into place. So I found somewhere on the net the definition n! = (1-1/2)×(1-1/3)²... × (1-1/n)^n-1] × nⁿ.

I will then use this result and the definition of e^x since we are talking for large n to proceed.

Homework Statement

Problem is, how do I prove it? Although I haven't tried induction yet, I was told that I should find an algebraic method.

I don't want a complete solution, just a hint to guide me through the proof.

Thank you

 

[Bonaparte]

First represent all the 1-.5 and 1-1/3 by (1-1/n+1)^n, we note this is (n/(n+1))^n.
Now note what happens as we multiply this for different n's:
(1/2)^1 * (2/3)^2 * (3/4)^3... = (1^1 * 2^2 * 3^3...)/(2^1 * 3^2 * 4^3...).
Do you notice what is happening? Try canceling things up, in problems like this, I recommend you put things in variables like I did (n/(n+1)) in our case.Thanks, Bonaparte

 

[Ryuky]

First represent all the 1-.5 and 1-1/3 by (1-1/n+1)^n, we note this is (n/(n+1))^n.
Now note what happens as we multiply this for different n's:
(1/2)^1 * (2/3)^2 * (3/4)^3... = (1^1 * 2^2 * 3^3...)/(2^1 * 3^2 * 4^3...).
Do you notice what is happening? Try canceling things up, in problems like this, I recommend you put things in variables like I did (n/(n+1))^n in our case.Thanks, Bonaparte

Thank you for your response. I got it now, the fraction reduces to n!/(n+1)^n = Product from 1 to n of (n/(n+1))^n which leads to n!=(n+1)^n * the Product, but then the last term of the product cancels out with (n+1)^n and so comes the required result.

However, I am still interested on how you derived it. I mean, suppose we had absolutely no clue on how does n^n relate to n!. How would we then, think of this relationship? What steps would we have to take?

 

[Bonaparte]

The relationship between n^n and n! is ruled by their definition. Now that I think about it a clearer way to have stated this is instead of n/n+1 (where we have to change n^n to (n+1)^n) is to state it in their way (1-1/n)^(n-1).
Then we have ((n-1)/n)^(n-1). Then proceed as before, although now it will be clearer. Considering your writing a paper about it, I recommend the second way.

As for your question, I derived it by look at the numbers, seeing the pattern (which is not even necessary considering its given) and implementing it into numbers.

Thanks and good luck, Bonaparte