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

Attempt:

$$n!=n\left( n-1 \right)\left( n-2 \right)...\left( n-\left( n-1 \right) \right)=n^{n}+...$$

$$\lim_{n \to \infty} \frac{n!}{n^{n}}=\lim_{n \to \infty} \frac{n^{n}+...}{n^{n}}=\lim_{n \to \infty} \frac{1+...}{1}=1+\lim_{n \to \infty} \ ...=1$$

I based this on the fact that $n$ is the highest power. The answer's meant to be $0$ though.

jimjim
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user322637
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  • Well, the fact is that the coefficients of that polynomial depend on $n$, so you cannot just cancel them out. –  Mar 14 '16 at 06:37
  • i did polynomial expansion and i assumed (n-1) at the end to be some number. idk does that answer ur question ? – user322637 Mar 14 '16 at 06:37
  • @G.Sassatelli i depended on the powers not the coefficients – user322637 Mar 14 '16 at 06:38
  • I did something like this $n!=n^{n}+an^{n-1}+bn^{n-2}+cn^{n-3}+...$ – user322637 Mar 14 '16 at 06:44
  • I understand that. But the fact is that, for instance $$a=-\sum_{k=1}^{n-1}k=\frac{n-n^2}{2}$$ so $$an^{n-1}=\frac{n^n-n^{n+1}}{2}$$ which cannot be thrown away like you did. The same goes for the others, with more complicated furmulas. –  Mar 14 '16 at 06:45
  • @G.Sassatelli no idea what u did (I'm not that advanced in calculus yet), I only know that $n!=n\left( n-1 \right)\left( n-2 \right)...\left( n-b \right),; b=n-1$ which looks like a polynomial with highest power $n$. Sorry for my lack of experience – user322637 Mar 14 '16 at 06:50
  • Consider the polynomials $$p_k(x)=x(x-1) \cdots (x-k+1)=x^k+a^{(k)}{k-1}x^{n-1}+a^{(k)}{k-2}x^{k-2}+ \cdots +a^{(k)}1x$$ What you are stating is $n!=p_n(n)$. However, it turns out that, for instance, the coefficient $a^{(k)}{k-1}$ goes to $-\infty$ as $k\to\infty$ roughly like $-\dfrac{k^2}{2}$. So you cannot approximate $p_{n}(n)$ with just the highest degree monomial, because actually $$\left\lvert a^{(n)}_{n-1}n^{n-1}\right\rvert\approx \frac12n^{n+1}\gg n^n$$ –  Mar 14 '16 at 08:41

2 Answers2

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Here is an elementary way using Sandwich Theorem:

$$0 \leq \lim_{n \rightarrow \infty} \frac{n!}{n^n} = \lim_{n \rightarrow \infty} \frac {n(n-1)...3.2.1}{n^n} \leq \lim_{n \rightarrow \infty} \frac{1}{n} = 0$$

Edit: $$ \frac{n!}{n^n} = \frac{n \times (n - 1) \times \cdots \times 1}{n \times n \times \cdots \times n} = 1 \times \frac{n - 1}{n} \times \cdots \times \frac{1}{n} \leq 1 \times 1 \times \cdots \times \frac{1}{n} = \frac{1}{n}.$$

Arpit Kansal
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1

You can use series, for example:

$$a_n:=\frac{n!}{n^n}\implies\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1\implies$$

$$\implies\text{the series}\;\;\sum_{n=1}^\infty a_n\;\;\;\text{converges}\;\implies \lim_{n\to\infty}a_n=0$$

DonAntonio
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