- #1

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Here is my write-up. I apologize for not using the built-in equation editor.

http://i.imgur.com/iyAto.png

- Thread starter scorpion4377
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- #1

- 9

- 0

Here is my write-up. I apologize for not using the built-in equation editor.

http://i.imgur.com/iyAto.png

- #2

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[tex]\log(n!)=\sum_{k=1}^{n}\log(k) \sim \int_{1}^{n}\log(x)dx[/tex]

The integral on the right side can be evaluated using integration by parts:

[tex]\int \log(x)dx = x\log(x) - \int dx = x\log(x)-x+C[/tex]

and this yields

[tex]\log(n!) \sim n\log(n)-n \sim n\log(n)[/tex]

Can you take it from there?

- #3

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That wouldn't be rigorous, would it? In my proof, I kept the error terms and treated them rigorously.

EDIT: The point of the problem is to proof the asymptotic behavior of the function, so it would be self defeating to use its asymptotic form. Please correct me if I am wrong, though.

EDIT: The point of the problem is to proof the asymptotic behavior of the function, so it would be self defeating to use its asymptotic form. Please correct me if I am wrong, though.

Last edited:

- #4

AlephZero

Science Advisor

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$$\log(n!) = \sum_{k=2}^n \log k$$

I changed the lower limit since ##\log 1 = 0##.

Because ##\log## is a monotonic function you can bracket the sum by two integrals

$$\int_2^n \log (x-1)\,dx < \sum_{k=2}^n \log k < \int_2^n \log x\,dx$$

Draw a picture of the integrals and the sum as a set of rectangles, if you can't see where that came from.

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