Proof that sequence $\lim_{n \to \infty} \frac{n!}{n^{n}} = 0$ - explanation needed I'm working through Spivak's Calculus and a question asks to verify that sequence $$\lim_{n \to \infty} \frac{n!}{n^{n}} = 0$$.
I've been given the hint that $n! = n(n-1)\dots k!$ for $k < n$, in particular for $k < \frac{n}{2}$.
I didn't  succeed in getting a formal solution, but I have access to the solution manual and they did the following:
$$\frac{n!}{n^{n}} = \frac{n(n-1)\dots(\frac{n}{2})!}{n^{\frac{n}{2}}n^{\frac{n}{2}}} \leq \frac{(\frac{n}{2})!}{n^{\frac{n}{2}}} \leq \bigg(\frac{1}{2}\bigg)^{\frac{n}{2}}$$.
I have a few issues with the solution:
1) How is $n(n-1)\dots(\frac{n}{2})! < (\frac{n}{2})!$ ? This makes no sense to me considering that the left side is a larger factorial.
2) How does the relationship $$\frac{(\frac{n}{2})!}{n^{\frac{n}{2}}} \leq \bigg(\frac{1}{2}\bigg)^{\frac{n}{2}}$$
actually come about ?. I agree with it in theory, and I've seen similar expressions in previous work, but I haven't formally encountered this in the text (particularly with treating the factorial). There is probably a similar version to this somewhere on the site that I could look at and attempt to prove.
EDIT
I had forgotten to mention that I had rewritten the expression as
$$\frac{n(n-1)(n-2)\dots(n-k) \dots  1}{n \times n \times n \dots \times n} = \frac{n-1}{n} \frac{n-2}{n} \dots \frac{1}{n}$$
and if I took the limit as $n \to \infty$ for each term I could get $0$, but I felt that this was not the right thing to do and if there was a more formal approach.
 A: It’s easiest to see what’s going on in that calculation when $n$ is even. Suppose that $n=2m$. Then
$$\frac{n(n-1)\ldots(m+1)}{n^m}=\frac{n}n\cdot\frac{n-1}n\cdot\ldots\cdot\frac{m+1}n\le 1\,,$$
so
$$\frac{n!}{n^n}=\frac{n(n-1)\ldots(m+1)m!}{n^mn^m}\le\frac{m!}{n^m}\,.$$
And
$$\frac{m!}{n^m}=\frac{m}n\cdot\frac{m-1}n\cdot\ldots\cdot\frac1n\le\left(\frac12\right)^m\,,$$
because $\frac{k}n\le\frac12$ for $k=1,2,\ldots,m$.
If $n=2m+1$, you can write
$$\frac{n!}{n^n}=\frac{n(n-1)\ldots(m+1)}{n^{m+1}}\cdot\frac{m!}{n^m}\le\frac{m!}{n^m}\le\left(\frac12\right)^m$$
and use the same reasoning.
A: A different way
$$
\frac{n!}{n^{n}}  = \frac{1}{n}\frac{2}{n}\cdots\frac{n-1}{n}\frac{n}{n}
$$
Now for every positive number $k < n$ we have that
$$
\frac{n - k}{n} \le \frac{n-1}{n} = q_n < 1
$$
It is clear that for every $n$ and every power of $q_n$ we have $q^m_n < 1$. It folows that
$$
\frac{n!}{n^{n}}  = \frac{1}{n}\frac{2}{n}\cdots\frac{n-1}{n}\frac{n}{n} \le \frac{1}{n}\ \underbrace{q_n \cdots q_n}_{n-2 \text{ factor}} \ \frac{n}{n} = \frac{1}{n}\ q_n^{n-2} < \frac{1}{n}\ 
$$
Take the limits.
A: Observe that
\begin{align}
\frac{n!}{n^n}
=&\exp(\log n!-\log n^n)\\
=&\exp\left(\sum_{j=1}^n(\log j-\log n)\right)\\
=&\exp\left(\sum_{j=1}^n\log\frac jn\right)\\
=&\exp\left(n\sum_{j=1}^n\frac1n\log\frac jn\right)\\
\end{align}
and since
$$
\sum_{j=1}^n\frac1n\log\frac jn\longrightarrow\int_0^1\log x dx<0
$$
you can conclude.
