# Stirling's approximation limit problem

1. May 26, 2009

### nitai108

Hi, I don't undestand this limits to infinity.

[ (2n)! / (n!)^2 ]^1/n

and

[ ( n^n + 2^n ) / (n! + 3^n) ]^1/n

I've absolutely no idea how the first one can be "4", and the seconda one "e". Assuming (1 + 1/n)^n = e I don't see how can I go from that form to this one. Any hint is appreciated.

2. May 26, 2009

### tiny-tim

Welcome to PF!

Hi nitai108! Welcome to PF!

(try using the X2 tag just above the Reply box )
Hint: what, approximately, is (2n)! / n! ?
Hint: which bits of the top and the bottom can you ignore?

3. May 26, 2009

### Random Variable

Re: Limits

To evaluate the first limit I would use Stirling's approximation, namely that $$n! \approx n^{n}e^{-n}$$ (and $$(2n)! \approx (2n)^{2n}e^{-2n}$$) for large n.

Last edited: May 27, 2009
4. May 27, 2009

### Random Variable

Re: Limits

For the second limit, do what tiny-tim suggested and then use Stirling's approximation.

5. May 28, 2009

### nitai108

Re: Limits

Thanks tiny-tim, and thanks Random Variable, I did solve them pretty easily with the Stirling's approximation.

Is there any other way to solve them, without Stirling's approx.?

6. May 28, 2009

### tiny-tim

The first one, you could say (2n)! is approximately (2n)2(2n-2)2 … 22

(but you'd have to prove that rigorously)

… which makes it fairly easy

7. May 29, 2009

### nitai108

Re: Limits

Ok, thanks a lot.

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