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This question already has an answer here:

I have to prove that

$$\lim_{n\to \infty} \frac {n^n} {n!}=\infty$$

I've tried to look for a lower bound that also converges to $\infty$ (I don't know if I'm explainig myself correctly), but I haven't found one yet. Applying L'Hôpital is way too complicated in $n!$, and the epsilon proof does not work as I have no way whatsoever of finding N.

Any ideas?

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marked as duplicate by MJD, Thomas, Daniel Fischer, froggie, egreg Feb 14 '14 at 20:23

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    $\begingroup$ ${n^n\over n!}=\color{maroon}{{n\over\vphantom{1} n}{n\over n-1}\cdots {n\over 2}}{n\over 1}\ge \color{maroon}1\cdot n$. $\endgroup$ – David Mitra Feb 14 '14 at 18:48
  • $\begingroup$ intuitively, $$\frac{n^n}{n!}=\frac{n\cdot{n}\cdot{...}\cdot{n}\cdot n}{n(n-1)...2\cdot{1}}$$ $\endgroup$ – Eleven-Eleven Feb 14 '14 at 18:49
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    $\begingroup$ Alternatively you can prove that $\frac{n!}{n^n} \to 0$, hence the reciprocal tends to $\infty$ $\endgroup$ – Alex Feb 14 '14 at 18:51
  • $\begingroup$ @Alex that´s the limit I wanted in the first place, but for some reason I thought this was easier. $\endgroup$ – Lessa121 Feb 14 '14 at 18:52
  • $\begingroup$ @DavidMitra Thanks! $\endgroup$ – Lessa121 Feb 14 '14 at 18:53
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Setting $$ a_n=\frac{n^n}{n!}, $$ we have $$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\frac{(n+1)^n\cdot n!}{n!\cdot n^n}=\frac{(n+1)^n}{n^n}=\left(1+\frac{1}{n}\right)^n \quad \forall n. $$ Since $$ \lim_n\frac{a_{n+1}}{a_n}=\lim_n\left(1+\frac{1}{n}\right)^n=e>2, $$ there is an $N \in \mathbb{N}$ such that $$ \frac{a_{n+1}}{a_n}>2 \quad \forall n\ge N. $$ It follows that $$ a_n\ge 2^{n-N}a_N \quad \forall n\ge N, $$ thus $$ \lim_na_n\ge \lim_n2^{n-N}a_N, $$ i.e. $\lim_na_n=\infty$.

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check this out:

http://en.wikipedia.org/wiki/Stirling%27s_approximation

This is a famous estimation of n!.

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