3
$\begingroup$

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.

Thanks!

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

$\endgroup$
  • 5
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.