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Evaluate the limit $$\lim_{n\to \infty}{\frac{(n+3)!}{n^n}}, n\in \mathbb N$$ I know that the limit is $0$ but how to prove it?

Martin Sleziak
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Breldor
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  • Use the Stirling's approximation. –  Nov 09 '14 at 20:04
  • It is $0$ because $n^{n}$ goes to infinity faster than $(n+3)!$ – Dipok Nov 09 '14 at 20:07
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    No need to use such heavy weapons. This has an elementary proof. – user2345215 Nov 09 '14 at 20:07
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    Dipok but how do you prove it? – Breldor Nov 09 '14 at 20:08
  • Very similar questions: http://math.stackexchange.com/questions/61713/whats-the-limit-of-the-sequence-lim-n-rightarrow-infty-fracnnn/61741#61741, http://math.stackexchange.com/questions/535226/limit-of-the-sequence-nn-n, http://math.stackexchange.com/questions/397866/limits-of-sequences-exponential-and-factorial, http://math.stackexchange.com/questions/877305/limit-of-factorial-function-lim-limits-n-to-infty-fracnnn You can find several relatively simple solutions there. Very probably similar ideas will work here. – Martin Sleziak May 20 '15 at 09:48

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Note that $$\frac{(n+3)!}{n^n}=2\cdot3\cdot\frac{4\cdot\ldots\cdot n\cdot(n+1)\cdot(n+2)\cdot(n+3)}{n\cdot\ldots\cdot n}\le 6\cdot\frac4n\cdot\frac{n^{n-4}}{n^{n-4}}\cdot\frac{(2n)^3}{n^3}=6\cdot\frac4n\cdot8$$

user2345215
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