How to find the limit of the sequence $a_n=\frac{n^n}{3^n\cdot n!}$ as $n$ tends to infinity.

Question

Let $a_n=\dfrac{n^n}{3^n\cdot n!}.$ Show that $a_n\to0$ as $n\to\infty$.

I know that I can use the ratio test for sequences, and $n^n$ increases faster than $3^n \cdot n!$ so it will tend towards infinity so I invert the sequence so I have to show that $\frac{3^n \cdot n!}{n^n}$ tends towards $0$.

I divide $a_{n+1}$ by $a_n$, I get $$\frac{3^{n+1} \cdot (n+1)! \cdot n^n}{(n+1)^{n+1} \cdot 3^n \cdot n!}.$$ I can get simplify up to get $\frac{3n^n}{(n+1)^{n+1}}$ but I do not know how to simplify any further so I can find the limit as $n$ tends to infinity.

Answer 1

You have a small error: the ratio simplifies to $$\frac{(n+1)^n}{3\,n^n}.$$

Hint:

$$\lim_{n\to\infty}\Bigl(1+\dfrac1n\Bigr)^n=\mathrm e.$$

Answer 2

Proof

Consider the positive series $\sum\limits_{n=1}^{\infty}a_n.$ Notice $$\frac{a_{n+1}}{a_{n}}=\frac{1}{3}\left(1+\frac{1}{n}\right)^n.$$It's well-known that $\left(1+\dfrac{1}{n}\right)^n$ increases with an increasing $n$. Hence $$\left(1+\frac{1}{n}\right)^n<\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e<3，$$which implies that $$\frac{a_{n+1}}{a_n}<1.$$By the ratio test, we may claim $\sum\limits_{n=1}^{\infty}a_n$ is convergent. Therefore $$\lim_{n \to \infty}a_n=0,$$ which is the necessary condition for a convergent seriers.

Answer 3

Another Proof

By Stirling's formula, we have

$$n! \sim \sqrt{2\pi n}\cdot\frac{n^n}{e^n},~~~n \to \infty$$

Thus, $$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\dfrac{n^n}{3^n\cdot n!}=\lim_{n \to \infty}\dfrac{n^n}{3^n\cdot \sqrt{2\pi n}\cdot\dfrac{n^n}{e^n}}=\lim_{n \to \infty}\left[\left(\frac{e}{3}\right)^n\cdot \frac{1}{\sqrt{2\pi n}}\right]=0\cdot 0=0.$$

Answer 4

Correct if wrong.

Second attempt:

$n \in \mathbb{Z^+}$.

$$e^n= 1+ n +n^2/2! ....+n^n/n!+ n^{n+1}/(n+1)!+......,$$

implies $e^n > (n^n/n!) $, $n \in \mathbb{Z^+}$.

$\dfrac{e^n}{3^n} > \dfrac{n^n}{3^n n!} >0$.

Note: $e/3 < 1$ ,

hence $\lim_{n \rightarrow \infty} (e/3)^n =0.$

Squeeze:

$\lim_{n \rightarrow \infty} (e/3)^n \ge \lim_{ n \rightarrow \infty} a_n \ge 0.$

Answer 5

A Third Proof

Since $$\frac{a_{n}}{a_{n-1}}=\frac{1}{3}\left(1+\frac{1}{n-1}\right)^{n-1}$$ for $n=2,3\cdots$

then $$a_{n}=a_1\cdot\prod_{k=2}^{k=n}\frac{a_{k}}{a_{k-1}}=a_1\cdot \prod_{k=2}^{k=n}\frac{1}{3}\left(1+\frac{1}{k-1}\right)^{k-1}.$$

Notice that $$\left(1+\frac{1}{k-1}\right)^{k-1}<e$$ for $k=2,3,\cdots$

Thus, $$0<a_{n}<\frac{1}{3} \cdot \left(\frac{e}{3}\right)^{n-1}$$for $n=2,3\cdots.$

Let $n \to \infty$. Since $0<\dfrac{e}{3}<1$,then $\left(\dfrac{e}{3}\right)^{n-1} \to 0$. By the squeeze theorem, we may claim $$\lim_{n \to \infty}a_n=0.$$