# What is the limit of the sequence n!/4^n? [duplicate]

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).

• why do you think the answer should be zero? can you show us your attempt?
– DHMO
Apr 16, 2017 at 9:39
• $\displaystyle \lim_{n\to\infty} \dfrac{n!}{4^n} = \dfrac14 \cdot \dfrac24 \cdot \dfrac34 \cdot \dfrac44 \cdot \dfrac54 \cdots = \infty$
– DHMO
Apr 16, 2017 at 9:40
• Note that $\displaystyle \limsup_{n\to\infty} \sqrt[n]{n!} = \infty$
– DHMO
Apr 16, 2017 at 9:42
• See also Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. and other questions linked there. Apr 16, 2017 at 15:11
• why lim sup =inf ??
– Owen
Apr 16, 2017 at 15:42

A fancy and nice (imo) way to do it. Take

$$a_n:=\frac{4^n}{n!}\implies\frac{a_{n+1}}{a_n}=\frac{4^{n+1}}{(n+1)!}\frac{n!}{4^n}=\frac4{n+1}\xrightarrow[n\to\infty]{}0<1\implies \sum_{n=1}^\infty a_n$$

$$\text{converges}\;\implies \lim_{n\to\infty}a_n=0\implies\;\lim_{n\to\infty}\frac1{a_n}=\infty$$

• The first expression you wrote for $a_n$ is incorrect. Apr 16, 2017 at 10:04
• I like this approach more than any other.+1 Apr 16, 2017 at 10:05
• @HocNgo I don't understand: what is wrong? Apr 16, 2017 at 11:07
• @DonAntonio, you swapped the numerator and denominator. The correct expression for $a_n$ should be $$a_n = \frac{n!}{4^n}.$$Hmm, I just realized your $a_n$ is not the same as the one I thought in the original problem. My apology. Apr 16, 2017 at 11:13
• @HocNgo At the end, he considers $1/a_n$, which is your limit. Apr 16, 2017 at 11:23

One may note that

$$n!=1\times2\times3\times4\times5\times\dots\times n>24\times5^{n-4}$$

Thus,

$$a_n>\frac{24}{5^4}\times\left(\frac54\right)^n\to\infty$$

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

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 $\displaystyle{\lim_{n\to\infty}u_n=+\infty}$

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 $1$.

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.

• why n!=exp(n*ln n -n)??
– Owen
Apr 16, 2017 at 15:21
• – DHMO
Apr 16, 2017 at 15:22
• Perhaps it is worth adding a link to this question: $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite, where various approaches to the limit of $\sqrt[n]{n!}$ can be seen in the answers. Apr 16, 2017 at 16:01