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Stirling's approximation limit problem

  • Thread starter nitai108
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi nitai108! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
[ (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? :wink:
 
  • #3


To evaluate the first limit I would use Stirling's approximation, namely that [tex] n! \approx n^{n}e^{-n} [/tex] (and [tex] (2n)! \approx (2n)^{2n}e^{-2n} [/tex]) for large n.
 
Last edited:
  • #4


For the second limit, do what tiny-tim suggested and then use Stirling's approximation.
 
  • #5
14
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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
tiny-tim
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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 :wink:
 
  • #7
14
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Ok, thanks a lot.
 

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