# finding limit Squeeze theorem [closed]

use $\lim_{n\to \infty} (1-\frac{1}{n})^n = \frac{1}{e}$

together with squeeze theorem to find $lim_{n\to \infty}\frac{n!}{n^n}$

I dont know how to use the first part

## closed as off-topic by Davide Giraudo, Claude Leibovici, Watson, Daniel W. Farlow, Jack's wasted lifeOct 4 '16 at 0:53

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• Look at the ratio $\dfrac{n!/n^n}{(n-1)!/(n-1)^{n-1}}$. – Daniel Fischer Oct 2 '16 at 17:44
• this is from a chapter before ratio test is introduced. I'm supposed to connect squeeze theorem to the first part – 345 Oct 2 '16 at 17:49
• @DanielFischer As you're a moderator, I'd like to ask you a question. Would it be acceptable to write a hint like yours as an answer? I see a lot of people writing hints in the answers. Why didn't you write yours as one? Edit: I see that your hint doesn't use the squeeze theorem although the poster asked for that specifically. Maybe this is the reason why? – Skeleton Bow Oct 2 '16 at 18:07
• @SkeletonBow Opinions on hints differ. I don't consider the above hint to be an appropriate answer, hence I didn't post it as one. My hint was intended to lead to an application of the squeeze theorem. – Daniel Fischer Oct 2 '16 at 18:28

We present two ways forward. The first follows the suggestion in the comment left by @danielfischer while the second bypasses the use of the limit for $e^{-1}$.

METHODOLOGY $1$:

Let $a_n=\frac{n!}{n^n}$. Then, the ratio $\frac{a_n}{a_{n-1}}$ can be written for $n\ge 2$

$$\frac{a_n}{a_{n-1}}=\left(1-\frac1n\right)^{n-1} \tag 1$$

In THIS ANSWER, I showed that the sequence $\left(1-\frac1n\right)^n$ is monotonically increasing. Inasmuch as it converges to $e^{-1}$, we see from $(1)$ that

$$\frac{a_n}{a_{n-1}}\le \frac{ne^{-1}}{n-1} \tag 2$$

Using $(2)$ recursively reveals that

\begin{align} 0\le a_n=\frac{a_n}{a_{n-1}}\,\frac{a_{n-1}}{a_{n-2}}\,\frac{a_{n-2}}{a_{n-3}}\cdots \frac{a_3}{a_{2}}\,\frac{a_2}{a_{1}}\,a_1 \le ne^{-{n-1}} \tag 3 \end{align}

whereupon applying the squeeze theorem to $(3)$ yields the limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{n!}{n^n}=0}$$

METHODOLOGY $2$:

We can write the ratio $\frac{n!}{n^n}$ as

\begin{align} 0\le \frac{n!}{n^n}&=\underbrace{\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{n-2}n\right)}_{\text{All terms are less than}\,\,1}\,\,\underbrace{\left(1-\frac{n-1}n\right)}_{=1/n}\le \frac1n\tag 4 \end{align}

whereupon applying the squeeze theorem to $(4)$ yields the limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{n!}{n^n}=0}$$