# Stirling's approximation limit problem

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
Homework Helper
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? 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.

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

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
Homework Helper
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 Ok, thanks a lot.