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Well I can Intuitively see that.

I am wondering If there is a neat way to prove that

Asaf Karagila
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Holy cow
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    Please make the body of the question self contained. When you read a book you don't have the first chapter starts straight from the title, do you? – Asaf Karagila Jul 02 '14 at 03:49
  • Stirling's approximation? – T.J. Gaffney Jul 02 '14 at 03:54
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    See this, but replace $a$ with $x$: http://math.stackexchange.com/questions/849084/proving-a-sequence-converges – JimmyK4542 Jul 02 '14 at 03:55
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    @JimmyK4542: My advisor tells a cute story that when he was in grad school (or fresh out of it, can't remember) he taught basic calculus to medical students (or something equivalent in level). And they copied from one another all the time. In one of the exercises they had a limit whose limit involved $e$. Trying to be clever people who copied changed letters of variables and functions. And many changed $e$ as well. Some of the papers had a foot note "We take $a=e$". :-) – Asaf Karagila Jul 02 '14 at 03:59
  • I have a meta question: This question has been marked as duplicate. However, my answer below exists nowhere in the duplicate thread. Would it be appropriate for me to cross-post it over there? – Kaj Hansen Jul 02 '14 at 05:31
  • @Kaj Hansen, I guess you should – Holy cow Jul 02 '14 at 05:53
  • Oh never mind. It actually is in the other thread...didn't see it at first. – Kaj Hansen Jul 02 '14 at 08:08

4 Answers4

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We can apply the ratio test to the sum $\displaystyle \sum \frac{x^n}{n!}$.

$$\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n} = \lim_{n \rightarrow \infty} \Bigg(\frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^{n}} \Bigg) = \lim_{n \rightarrow \infty} \Big(\frac{x}{n+1}\Big)$$

This limit goes to zero for all $x \in \mathbb{R}$. We can conclude that the series converges, and so....

Kaj Hansen
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Stirling's Approximation states that: $$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$ The rest should be rather obvious.

Nathaniel Bubis
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Let $m$ be the smallest integer which is $\ge 2|x|$. Let $B=\frac{|x|^m}{m!}$.

It is possible that $B$ is fairly big. However, when we continue past $m$, each time we increment $n$ by $1$, our expression goes down in absolute value by a factor of at least $2$. Thus $$\frac{|x|^{m+t}}{(m+t)!}\lt B \frac{1}{2^t}.\tag{1}$$ As $t\to\infty$, the right-hand side of (1) approaches $0$.

Remark: Probably not neat, nor cool. But it makes precise the intuition that after a while, increasing $n$ makes a much bigger contribution to the bottom than to the top, so after a while poor $x^n$ can't keep up with the factorial.

André Nicolas
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I more intuitive answer.

$$\lim\limits_{n\to\infty}\frac{x^n}{n!}=\frac{x.x.x.\dots}{1.2.3.\dots}$$.

For some integer $N$, we have $N\leq x<N+1$. Hence, $\lim\limits_{n\to\infty}\frac{x^n}{n!}=\frac{x.x.x.\dots}{1.2.3.\dots}$ is the product of a finite number of real numbers greater than $1$, and an infinite number of real numbers less than $1$. Moreover, the list of real numbers less than $1$ is a strictly decreasing sequence.

So say the product $\frac{x.x.x\dots x}{1.2\dots N}=a$. Then $\lim\limits_{n\to\infty}\frac{x^n}{n!}<a.\left(\frac{x}{N+1}\right)^{n-N}$.

We know that $\frac{x}{N+1}<1$

The limit has to be $0$.