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I have a homework question, so please don't answer fully but I would appreciate a push in the right direction.

Basically we need to figure out if $n^{n+\frac{1}{2}}e^{-n}$ is larger,smaller, or equal to $n!$ as $n$ goes to infinity.

I checked the answer, and it turns out they are equal. when $n$ goes to infinity, one divided by the other does not yield $0$. But how do I show it?

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  • $\begingroup$ Are you familiar with the Laplace method? $\endgroup$ – Antonio Vargas Dec 16 '13 at 15:25
  • $\begingroup$ No. I've never heard or was taught the Laplace method. $\endgroup$ – Oria Gruber Dec 16 '13 at 15:26
  • $\begingroup$ And you haven't been shown Stirling's formula either? $\endgroup$ – Antonio Vargas Dec 16 '13 at 15:27
  • $\begingroup$ this may be useful: en.m.wikipedia.org/wiki/Stirling's_approximation $\endgroup$ – mojambo Dec 16 '13 at 15:28
  • $\begingroup$ We have. Is stirling approximation the way to solve this? $\endgroup$ – Oria Gruber Dec 16 '13 at 15:28
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My suggestion is to take logs of both sides, and for $n!$ consider the integral of $\log x.$

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  • $\begingroup$ Thanks for the great advice. But as said in the comments, I did manage to solve it using stirling approximation. Will accept answer to close this page. Thank you. $\endgroup$ – Oria Gruber Dec 16 '13 at 15:49
  • $\begingroup$ @OriaGruber Doing it this way is not better, but slightly simpler than Stirling (you are actually proving a form of Stirling approx.) $\endgroup$ – Igor Rivin Dec 16 '13 at 15:51

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