What is the limit of the sequence n!/4^n? I am trying to find the limit of the sequence by using root test and I don't understand why the limit is not zero?
(the answer is inf).
 A: All the answers given so far use valid techniques for showing that the sequence goes to infinity. But don't forget you can always go back to the definition of a limit. The limit of a sequence is infinite if you can choose any value, call it $K$, and find an index $k$ such that $a_n > K$ for all $n > k$.  So, we can pick a value -- one billion -- and find a $k$ such that every $a_n$ past that is greater than a billion. 
Can you prove that? Hints: can you show that:


*

*once you are past the first few elements, the sequence is greater than one

*once you are past the first few elements, the sequence is monotone increasing

*that ${5 K}/{4} > K$ 

*that these three facts can be combined to show a proof of the desired property

A: Let $\displaystyle{u_n=\frac{n!}{4^n}}$. We observe that :
$$\frac{u_{n+1}}{u_n}=\frac{n+1}{4}\underset{n\to\infty}{\longrightarrow}+\infty$$
Hence, for $n\ge N$, with $N$ large enough :
$$\frac{u_{n+1}}{u_n}\ge2$$
and thus :
$$\forall n\ge N,\,u_n\ge 2^{n-N}u_N$$
This proves that $\displaystyle{\lim_{n\to\infty}u_n=+\infty}$
A: You can use Stirling's approximation for very large $n$ ,  $$n! \approx n^ne^{-n} \sqrt{2\pi n}$$
Your limit becomes : $$\displaystyle \lim_{n\to\infty} \dfrac{n!}{4^n} = \lim_{n \to \infty} \Big(\frac{n}{4e}\Big)^n\sqrt{2\pi n}=\infty $$
Since $n$ is very large, doesn't matter if you divide it by $4e$, it will also be very large than $1$.
A: By the root test:
$$\begin{array}{rcl}
\displaystyle \limsup_{n\to\infty} \sqrt[n]{a_n} &=& \displaystyle \limsup_{n\to\infty} \sqrt[n]{\dfrac{n!}{4^n}} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \sqrt[n]{n!} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \sqrt[n]{\exp\left(n \ln n - n\right)} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \exp\left(\ln n - 1\right) \\
&=& \displaystyle \dfrac1{4e} \limsup_{n\to\infty} n \\
&=& \infty
\end{array}$$
Hence the sequence diverges to infinity.
A: A fancy and nice (imo) way to do it. Take
$$a_n:=\frac{4^n}{n!}\implies\frac{a_{n+1}}{a_n}=\frac{4^{n+1}}{(n+1)!}\frac{n!}{4^n}=\frac4{n+1}\xrightarrow[n\to\infty]{}0<1\implies \sum_{n=1}^\infty a_n$$
$$\text{converges}\;\implies \lim_{n\to\infty}a_n=0\implies\;\lim_{n\to\infty}\frac1{a_n}=\infty$$
A: One may note that
$$n!=1\times2\times3\times4\times5\times\dots\times n>24\times5^{n-4}$$
Thus,
$$a_n>\frac{24}{5^4}\times\left(\frac54\right)^n\to\infty$$
