prove that $\frac{n!}{n^n}\le(\frac{1}{2})^k$ where $k=[\frac{n}{2}]$, the greatest integer $\le\frac{n}{2}$ 
1) For $n\ge 2$, prove that $\frac{n!}{n^n}\le(\frac{1}{2})^k$ where $k=[\frac{n}{2}]$, the greatest integer $\le\frac{n}{2}$.
  2) Deduce the value of $\underset{n\rightarrow\infty}{\lim}\frac{n!}{n^n}$.
  3) Show that the series $\underset{n=1}{\overset{\infty}{\sum}}\frac{n!}{n^n}$ is convergent and again deduce $\underset{n\rightarrow\infty}{\lim}\frac{n!}{n^n}$.

My attempt


*

*$\forall n\in\mathbb{N}$
$1<1+\frac{1}{n}\le 2\Rightarrow 1>\frac{1}{1+\frac{1}{n}}\ge \frac{1}{2}\Rightarrow 1>\frac{1}{(1+\frac{1}{n})^n}\ge \frac{1}{2^n}$
$\Rightarrow 1>\frac{a_{n+1}}{a_n}\ge\frac{1}{2^n}\Rightarrow a_n\ge a_{n+1}\ge\frac{a_n}{2^n}$
Using induction we eventually get, $a_1\ge a_n\ge\frac{a_1}{2^{\frac{n(n-1)}{2}}}\Rightarrow 1\ge a_n\ge\frac{1}{2^{\frac{n(n-1)}{2}}}$

*Using Sandwich theorem, we get $\underset{n\rightarrow\infty}{\lim}a_n=1$.


I couldn't get the required bound for $\frac{n!}{n^n}$ although, with or without using the result from 1 I can conclude the limit to be $1$.
  I did get an answer to my doubt at https://math.stackexchange.com/a/1226238/562589 but I couldn't understand how they concluded $\frac{k}{n}\frac{k-1}{n}...\frac{1}{n}\le(\frac{1}{2})^k$.


*Let $a_n=\frac{n!}{n^n}$.
Then, $\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=(\frac{n}{n+1})^n=\frac{1}{(1+\frac{1}{n})^n}$
$\Rightarrow\underset{n\rightarrow\infty}{\lim}\frac{a_{n+1}}{a_n}=\underset{n\rightarrow\infty}{\lim}\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}<1\Rightarrow\underset{n=1}{\overset{\infty}{\sum}}\frac{n!}{n^n}$ is convergent

I am not sure how to use this to deduce $\underset{n\rightarrow\infty}{\lim}\frac{n!}{n^n}$ differently(?) than done before.

 A: Q(2):
$$\dfrac{n!}{n^n}=\dfrac{1}{n^{n-1}}\times\dfrac{n-1}{n^n}\times...\times\dfrac{1}{n^n}$$
Apply L'Hopital's rule to each term, and Since every term in this expression can form a convergent sequence with limit equal to $0$. And:$$\lim_{n\to\infty}(s_nt_n)=\lim_{n\to\infty}(s_n)\lim_{n\to\infty}(t_n)$$ So:
$$\implies \lim_{n\to\infty}\dfrac{n!}{n^n}=0$$
Q(3)
You used the ratio test to conclude that $\sum(a_n)$ converges
We know that:$$\lim_{n\to\infty}\inf|\dfrac{a_{n+1}}{a_n}|\le\limsup|a_n^{1/n}|\le\limsup|\dfrac{a_{n+1}}{a_n}| (*)$$exists,because$\dfrac{a_{n+1}}{a_n}=\dfrac{1}{e}<\infty$.
So $(*)<\infty$.
$$\implies\lim(a_n)<\infty$$ Hence that limit exists, then we can evaluate that value an therefore prove that it is less than $[\frac{n}{2}]$.
A: If $n$ is even, the first $\frac n2$ factors of $$\frac {1\cdot 2 \cdot 3 \ldots n}{n^n}$$ are less than or equal to $\frac 12$ and the rest are less than or equal to $1$, so the fraction is less than or equal to $\left(\frac 12\right)^{n/2}$.  If $n$ is odd we just need to account for one more factor $\frac 12$.  Note that $\frac {3!}{3^3}=\frac 6{27}\lt \left(\frac 12\right)^2$ and for any $n$ higher than $3$ the first term is less than $\frac 14$ which gives one more factor $\frac 12$ and we are done.
