I need to prove that $$\lim_{n \rightarrow \infty} \frac {a^n} {n!}=0$$

I have no condition over $a$, just that is a real number. I thought of using L'Hôpital, but it's way too complicated for something that it should be simpler. Same goes to the epsilon proof, and I´m runnning out of options of what to do with it.


PD: I could also use a hint on how to solve $$\lim_{n \rightarrow \infty} \frac {n^n} {n!}=0$$

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    $\begingroup$ Hint Think about what happens when $n\gt a.$ $\endgroup$ – awllower Feb 14 '14 at 14:18
  • $\begingroup$ @awllower I did, but I can't see how that would help... $\endgroup$ – Lessa121 Feb 14 '14 at 14:50
  • $\begingroup$ I "formalized" it into an answer. Hope it helps then. :) $\endgroup$ – awllower Feb 14 '14 at 14:55

Compare it with the following geometric sequence:
$b_n=(\frac{a^m}{m!})(\frac{a}{m+1})^n,$ where $m$ is the smallest positive integer such that $m+1\gt a.$
Notice that $a_{n+m}\le b_n$ so that $\lim a_n=\lim b_n=0.$
Hope this helps.

  • $\begingroup$ The pleasure is mine. ;D $\endgroup$ – awllower Feb 14 '14 at 15:19
  • $\begingroup$ Only one thing, all of that is true only if n>a right? $\endgroup$ – Lessa121 Feb 14 '14 at 15:23
  • $\begingroup$ One way to express this is: when $n\gt a,$ then we can compare it to the geometric sequence defined above. Hence in response, I guess so. ;) $\endgroup$ – awllower Feb 14 '14 at 15:27

The result is clear if $|a|<1$. If $|a|>1$, I will now show the calculation with $a>1$ (it's no different with $a<-1$. The $n$-th value of the sequence equals $$\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdot\frac{a}{4}\cdots\frac{a}{n}.$$

Now, let $N$ be the integer which is larger than $a^2$. Let $C=\frac{a^N}{N!}$. Now take any integer $k$ and see that the $N+k$-th element of your sequence is $$C\cdot \frac{a}{N+1}\cdot\frac{a}{N+2}\cdots\frac{a}{N+k}\cdot$$ Since $\frac{a}{N+1}<\frac{a}{a^2}<\frac1a$, the $N+k$-th elements in the sequence is smaller than $$C\cdot \frac{1}{a^k},$$ meaning it goes to $0$.


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