2

Possible Duplicate:
How to prove that $\lim\limits_{n \to \infty} \frac{k^n}{n!} = 0$

Prove that for $a > 0$, $\lim_{n \rightarrow \infty}{\frac{a^n}{n!}=0}$. My attempt is since $$e^x=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$ and the series converges for all $x \in \mathbb{R}$, by the test of divergence, $$\lim_{n \rightarrow \infty}{\frac{a^n}{n!}=0}.$$ Is my proof correct ? or are there any alternative ?

Community
  • 1
Idonknow
  • 16,493
  • Yes, it is correct. – Siminore Feb 01 '13 at 15:40
  • It's correct, but uses a "sledgehammer" in my opinion. See this post and its links for more elementary methods. – David Mitra Feb 01 '13 at 15:41
  • This is also true for $a\leq 0$. – Julien Feb 01 '13 at 15:42
  • The proof is correct but, as @DavidMitra says, you're using a sledgehammer :) So I don't find it very clean, you should be able to prove then that those series are equal to $e^x$ and that they converge, using taylor's theorem also does the trick but it's more advanced that the analysis corresponding to this. Final message: I would look for another proof, but if I saw this in an exam and nothing came to my mind, I would undoubtedly use yours and it would be correct. – MyUserIsThis Feb 01 '13 at 15:49
  • @DavidMitra : What do you mean ' sledgehammer' in this case ? – Idonknow Feb 01 '13 at 15:53
  • By "sledgehammer', I mean that you're using a comparatively sophisticated result (the infinite series converges) to prove a result which can otherwise be proven using more elementary methods. This is fine; but a proof that uses only as much machinery as needed (in this case, all you need is that $a/n<1/2$ for $n$ sufficiently large and the comparision test for limits) would be preferable to some. – David Mitra Feb 01 '13 at 16:00

1 Answers1

1

In the interest of providing a (hopefully intuitive) alternative, since you asked for it:

By expanding the factorial (and thereby hopefully making it more tangible) it is fairly easy to see that it grows quicker than the exponential.

$$\frac{a^n}{n!} = \frac{a}{1}\frac{a}{2}\frac{a}{3}\ldots\frac{a}{n}$$

The exponential is simply the repeated product of $n$ number of $a$'s, whereas the factorial is the repeated product of $n$ ever-growing integers.

Therefore there must come a point, some $n \ge a$, after which all new factors of the denominator must be larger than the new factors of the numerator. This gap in the size of $a$ and $n$ will only continue to grow; the final factor itself, $\frac{a}{n}$, goes to $0$ as $n$ does, and so the product of an infinite amount of such small fractions has little choice but to follow suit and approach $0$.

Jakob Streipel
  • 1,859