What is the limit of the sequence n!/4^n?

Question

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).

Answer 1

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

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

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

Answer 2

One may note that

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

Thus,

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

Answer 3

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

Answer 4

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

Answer 5

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 $1$.

Answer 6

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.