How to show this limit approaches $0$?

Question

For the sequence $ a_n=\dfrac{2^n}{n!} $, is there a way to show that the $\lim_{n \to \infty }\dfrac{2^n}{n!}=0$?

I know that it is bounded above by $2$ because it is a decreasing sequence. I also know that the closest number it will every get to is $0$ because it won't be negative.

However, is there a way of showing that the limit goes to $0$?

Going from the $n$'th term to the $n+1$'st term, you are eventually more than halving it each time since the numerator grows by a factor of $2$ but the denominator grows by a factor of $n+1\gg 4$ — JMoravitz, Feb 27 '21 at 03:42
My favourite method: show $\sum \frac{2^n}{n!}$ is a convergent series by the ratio test, and conclude that the terms must therefore tend to $0$. — Theo Bendit, Feb 27 '21 at 04:15
Does this answer your question? Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. Found using Approach0. — Toby Mak, Feb 27 '21 at 06:30

score 2 · Answer 1 · answered Feb 27 '21 at 03:42

2

The trick is to see that for suiffiently large $n$, $n!>3^n.$

$$\lim_{n\to\infty}\frac{2^n}{n!}\leq\lim_{n\to\infty}\frac{2^n}{3^n} = \lim_{n\to\infty}\left(\frac{2}{3}\right)^n\to0.$$ See my answer here.

answered Feb 27 '21 at 03:42

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score 1 · Answer 2 · answered Feb 27 '21 at 03:53

OK. If you want to have a proof that may be different from those you've seen so far. Here is one. Observe that: $a_{n+1} = \dfrac{2^{n+1}}{(n+1)!}= \dfrac{2^n}{n!}\cdot \dfrac{2}{n+1}= \dfrac{2a_n}{n+1}$. With this recursive equation for the $a_n$'s coupled with $a_4 = \dfrac{16}{24} = \dfrac{2}{3} < 1$, you can by induction on $n \ge 4$ show that $a_n < 1$. Thus it follows that $a_n < \dfrac{2}{n}, n \ge 4$. So $a_n \to 0$ as desired by the squeeze lemma in analysis if you will.

score 1 · Answer 3 · answered Feb 27 '21 at 06:20

Intuition

$\frac {2^n}{n!} = \frac 21 \cdot \frac 22 \cdot \frac 23 \cdot \frac 24..... \cdot \frac 2n$. Surely as the terms of the product go to $0$ the product will go to $0$.

Well, intuition and \$2.75 will buy you a cup of coffee.

But if $a_n = \frac {2^n}{n!}$ then $a_1 = a_2 = 2$ and $a_{n+1}=a_n\cdot 2n$ and $\frac 2n < 1$ if $n \ge 3$ so for significantly large $n$ we have $a_n = (a_{n-1}) \cdot \frac 2n$ and $a_{n-1} < 2$ so $a_n < 2\cdot \frac 2n =\frac 4n$ and as $\frac 4n \to 0$ then be squeeze theorem $a_n \to 0$.

How to show this limit approaches $0$?

3 Answers3