How to solve : $\lim_{n\rightarrow \infty} \frac{n!}{n\cdot 2^{n}}$ $$\lim_{n\rightarrow \infty} \frac{n!}{n\cdot 2^{n}}$$
I need to solve the limit problem above. I have no idea about what to do. What do you suggest?
Thanks in advance.
 A: Since
$$n!=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \cdots \cdot n\ge 2\cdot 3\cdot 4^{n-3}$$
$$\frac{n!}{n\cdot 2^n} \ge 2\cdot 3\cdot \frac{4^{n-4}}{2^n}=\frac{2\cdot3}{4^4} 2^n$$
for $n\ge 4$. Therefore the sequence diverges.
A: Look at the series
$$\sum_{n=1}^\infty\frac{n2^n}{n!}=\sum_{n=1}^\infty a_n$$
and apply the quotient test
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)2^{n+1}}{(n+1)!}\cdot\frac{n!}{n2^n}=\frac2n\xrightarrow[n\to\infty]{}0$$
and thus the series converges, which means
$$a_n\xrightarrow[n\to\infty]{}0\implies\frac1{a_n}=\frac{n!}{n2^n}\xrightarrow[n\to\infty]{}\infty$$
A: Hint : 
$$\frac{n!}{n\cdot2^n} = \frac{(n - 1)!}{2^n}\\
= \frac{1}{2}\cdot\frac{(n-1)(n-2)...(3)(2)(1)}{\underbrace{2\cdot2\cdot2\cdot...\cdot2}_{\text{n-1 times}}}\\
= \frac{1}{2}\cdot\frac{n-1}{2}\cdot\frac{n-2}{2}\cdot\dots\cdot\frac{3}{2}\cdot\frac{1}{2}$$
Hopefully you can see how this is divergent.
A: Use Stirling's formula $n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n(1+O(\frac{1}{n}))$. You wil get
$\frac{n!}{n2^n}=\sqrt{\frac{2\pi}{n}}\left(\frac{n}{2e}\right)^n(1+O(\frac{1}{n}))\to  \infty$
