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

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.

• Please, use MathJax (i.e. LaTeX commands) for mathematical notations. Aug 4, 2018 at 9:42
• I agree with @user108128's assessment that this is a duplicate of the linked question, but the linked question is of quite poor quality, closed, and (now that it has been highlighted) somewhat likely to be deleted. This version of the question is a great improvement over the old version. I would like to suggest that this question be left open. Aug 5, 2018 at 1:41

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

Hint:

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

• Since I inverted the sequence originally as I have to show that an+1/an < 1 to show it is a null sequence and therefore the original sequence tends towards infinity, surely that it is 3*n^n/(n+1)^n and therefore it is 1/e and surely that is timesed by 3 as 3 is now on the numerator and 3/e>1 then my sequence isn't null but that isn't correct. Aug 4, 2018 at 10:51
• @M.Calculator: If $a_n$ is what is written at the beginning of your question, your simplification is wrong, and you obtain $\mathrm e/3$ as a limit of the ratio. Aug 4, 2018 at 11:31

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.

Another Proof

By Stirling's formula, we have

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

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

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

Thus, $$0for $$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 $$\lim_{n \to \infty}a_n=0.$$

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:

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

• Your answer is quite hard to read. Did you know that you can use  for displayed math (e.g. $$\frac{n!}{n^n} = \frac{n(n-1)\dotsc(n(n-1))}{n^n}$$ will by typeset as $$\frac{n!}{n^n} = \frac{n(n-1)\dotsc(n(n-1))}{n^n}.$$ You can also use the align environment for readability: \begin{align} A &= B \\ &= C \end{align} renders as \begin{align} A &= B \\ &= C. \end{align} Aug 5, 2018 at 1:49
• maybe you are wrong！how you get that$1(1-1/n)......(1-(n-1)/n) \gt (n-1)(1-(n-1)/n)=$? the left side is a product not a sum... Aug 5, 2018 at 3:32
• mangdie1982.Mistake. Thanks. Aug 5, 2018 at 7:18
• Xander Henderson.Thanks.In addition there is a mistake. Aug 5, 2018 at 7:22