Stirling's approximation limit problem

[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.

 

[tiny-tim]

Welcome to PF!

Hi nitai108! Welcome to PF!

(try using the X² tag just above the Reply box )

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

Hint: what, approximately, is (2n)! / n! ?

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

Hint: which bits of the top and the bottom can you ignore?

 

[Random Variable]

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.

 

[Random Variable]

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

 

[nitai108]

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.?

 

[tiny-tim]

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

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

 

[nitai108]

Ok, thanks a lot.