Why is $\lim_{n\to \infty} \frac{n^{4n}}{(4n)!} = 0$?

Question

I'm trying to figure out how to prove, that $$\lim_{n\to \infty} \frac{n^{4n}}{(4n)!} = 0$$ The problem is, that $$\lim_{n\to \infty} \frac{n^{n}}{n!} = \infty$$ and I have no idea how to prove the first limit equals $0$.

Answer 1

Using Stirling's approximation: $n! \approx c \sqrt{n}(n/e)^n$, we get $$\frac{n^{4n}}{(4n)!} \approx \frac{n^{4n}}{c\sqrt{4n}(4n/e)^{4n}} \approx \left(\frac{e}{4}\right)^{4n}\frac{1}{c\sqrt{4n}} \overbrace{\longrightarrow}^{\because \frac{e}{4}<1} 0.$$

Answer 2

The ratio of successive terms is

$$\frac{t_{n+1}}{t_n}=\frac{\dfrac{(n+1)^{4n+4}}{(4n+4)!}}{\dfrac{n^{4n}}{(4n)!}}=\left(\left(\frac{n+1}n\right)^n\right)^4\frac{(n+1)^4}{(4n+1)(4n+2)(4n+3)(4n+4)}.$$

The first factor is known to describe an increasing sequence that converges to $e^4$, while the second is asymptotic to $\dfrac1{256}$ and can be bounded above by $\dfrac1{105}$.

Finally,

$$\frac{t_{n+1}}{t_n}<\frac{e^4}{105}<1.$$

Answer 3

Define $$ \\a_n=\frac{n^{4n}}{(4n)!}=> \\\frac{a_{n+1}}{a_n}=\frac{(n+1)^{4(n+1)}}{n^{4n}}\cdot\frac{(4n)!}{(4n+4)!}= \\(n+1)^4\big(1+\frac{1}{n}\big)^{4n}\cdot\frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}=b_n $$ $$ \\\dfrac{(4n+1)(4n+2)(4n+3)(4n+4)}{(n+1)^4}=\Big(4-\frac{3}{n+1}\Big)\Big(4-\frac{2}{n+1}\Big)\Big(4-\frac{1}{n+1}\Big)\Big(4-\frac{0}{n+1}\Big)\ge(4-3/2)(4-2/2)(4-1/2)(4-0/2)=5\cdot6\cdot7\cdot8/2^4=105>3^4>((1+1/n)^n)^4=> \\b_n<81/105=>a_n<(81/105)^{n-1}=>\lim_{n\to+\infty}a_n=0 $$

Answer 4

Starting like other two answers: $$a_n = \frac{n^{4n}}{(4n)!}\implies \frac{a_{n+1}}{a_n} = \left(\left(1+\frac1n\right)^n\right)^4\cdot\frac{(n+1)^4}{(4n+1)(4n+2)(4n+3)(4n+4)}\to\frac{e^4}{4^4} < 1, $$ and by the quotient test $\sum a_n$ converges, but this implies $a_n\to 0$.