show that $\limsup(\frac{n!}{n^n})=0$

Let $a_n=\frac{n!}{n^n}$ be a sequence. Prove that $\limsup(a_n)=0$.

I know that the series $\sum^{\infty}_{n=1} \frac{n!}{n^n}$ converges, so the nth term must converge to 0. But I cannot use the convergence of series to prove this.

Basically what I have is: The sequence $a_n$ is decrescent and bounded by 0 $(a_n>0 \; \forall n)$, therefore is convergent and it's limit is the $\limsup(a_n)$.

So I must show that $\lim\frac{n!}{n^n}=0$

Let $\epsilon>0$, prove that exists $n_0 \in \mathbb{N}$ such that, if $n>n_0$

$$\left|\frac{n!}{n^n}-0\right|<\epsilon \Rightarrow -\epsilon<\frac{n!}{n^n}<\epsilon$$

I know that $\frac{n!}{n^n}>=-\epsilon$ because $\frac{n!}{n^n}>0$.

How can I prove that $\frac{n!}{n^n}<\epsilon$?

• Bound it by something you know goes to zero. Here's an idea: you have $\prod_{i=1}^n \frac{i}{n}$, note that all the terms in this product are less than or equal to $1$ and $\lfloor n/2 \rfloor$ of them are less than or equal to $1/2$. Can you conclude? – Ian Oct 8 '16 at 21:53
• How do you know the series converges? – πr8 Oct 8 '16 at 22:12

$$\dfrac{1}{n^n}\leq\dfrac{n!}{n^n}=\dfrac{n-1}{n}\dfrac{n-2}{n}\cdot\cdot\cdot\dfrac{1}{n}\leq1\cdot1\cdot 1\cdot\cdot\cdot\dfrac{1}{n}=\dfrac{1}{n}.$$

Does this help?

• +1 for you. Using this you can say $\frac{n!}{n^n} \leq \frac{1}{n} < \epsilon$. So pic $n_0 = \frac{1}{\epsilon}$ – Ahmad Bazzi Oct 8 '16 at 22:20

Hint $$\frac{n!}{n^n}=\frac{1}{n}\cdot \frac{2}{n} \cdot \frac{2}{n} \cdot...\cdot \frac{n}{n} \leq \frac{1}{n} \cdot 1 \cdot 1 \cdot ... \cdot 1$$

Another way more to show: using the quotient test for positive series, we get

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

so the series $\;\sum\limits_{n=1}^\infty\frac{n^n}{n!}\;$ converges, and thus

$$\lim_{n\to\infty}\frac{n^n}{n!}=0\implies\lim\sup_{n\to\infty}\frac{n^n}{n!}=\lim_{n\to\infty}\frac{n^n}{n!}=0$$

Suppose we choose any positive, sufficiently small $^{\mathbf{\dagger}}$ epsilon. We have $\displaystyle \frac{n!}{n^n} < \varepsilon \iff n! < \varepsilon (n^n)$.

Now taking the log of both sides and applying relevant properties, the above is true $\iff \displaystyle \sum_{k=1}^n \ln(k) < \ln(\varepsilon) + n \ln(n)$.

Rearranging yields $\displaystyle n \ln(n) - \sum_{k=1}^n \ln(k) \ = \ \sum_{k=1}^n \ln(n/k) > -\ln( \varepsilon)$.

We'll have $\displaystyle \left\lceil -\ln( \varepsilon) \right\rceil = M$ for some $M \in \mathbb{R}^+$. The above inequality holds for all $n > e^M$, so we may conclude $\displaystyle \lim_{n \rightarrow \infty} \left( \frac{n!}{n^n} \right) \rightarrow 0$.

$^{\mathbf{\dagger}}$ Important only so we don't have to worry about the sign of $\ln(\varepsilon)$.