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This question already has an answer here:

What I have got:

For any $a\in\mathbb{R}$, we can find an $N$, $N\gt a$, such that, $\lim_{n\to\infty}\frac{a^n}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdot\ldots\cdot\frac{a}{N-1}\cdot\frac{a}{N}\cdot\frac{a}{N+1}\cdot\ldots,$ and the later part $\frac{a}{N}\cdot\frac{a}{N+1}\cdot\ldots\lt\frac{a^m}{n^m}\to 0$ when $m\to\infty$, because $\frac{a}{n}\lt1$.

Then use the definition of convergence, with letting $M=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdot\ldots\cdot\frac{a}{N-1}$, and $\frac{a^m}{n^m}\lt\delta$ for $\delta=\frac{\epsilon}{M}$.

For an n that is large enough, we have $\frac{a^n}{n!}\lt\epsilon$, $\forall\epsilon\gt 0$.

What I want to ask is:

Are there any other ways to prove that this limit tends to $0$?

Thanks!

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marked as duplicate by Pedro Tamaroff, Dominic Michaelis, kahen, Martin Sleziak, Micah Apr 5 '13 at 17:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ i guess you totally aren't allowed to use stirlings approximation or ? $\endgroup$ – Dominic Michaelis Apr 5 '13 at 17:35
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    $\begingroup$ This has been asked over 9000 times here. Let me search. $\endgroup$ – Pedro Tamaroff Apr 5 '13 at 17:36
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    $\begingroup$ See this meta thread for FAQs on main and here for you question. $\endgroup$ – Pedro Tamaroff Apr 5 '13 at 17:38
  • $\begingroup$ i think i did not learn stirlings approximation... thank you i will see that thread. :) $\endgroup$ – chikurin Apr 5 '13 at 17:42
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We have for $a\neq 0$: $$\log\frac{|a|^n}{n!}=n\log| a|-\sum_{k=1}^n\log k\sim_\infty n\log|a|-\int_1^n\log x dx=n\log|a|-n\log n\to-\infty$$ hence $$\lim_{n\to\infty}\frac{a^n}{n!}=0$$

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You can use the result, if $ \lim_{n\to \infty} \frac{b_{n+1}}{b_n}=b $ and $|b|<1$, then $\lim_{n\to \infty} b_n = 0$. Applying this to your problem, we have

$$ \lim_{n \to \infty} \frac{a^{n+1}}{(n+1)!}\frac{n!}{a^n}=\lim_{n\to \infty}\frac{a}{n+1}=0 <1 $$

$$ \lim_{n \to \infty} \frac{a^n}{n!}=0. $$

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  • $\begingroup$ @user66607: You are welcome. $\endgroup$ – Mhenni Benghorbal Apr 6 '13 at 12:41
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If some asks a reasonable question is better to answer it straightforward than to elaborate into unnecessary details. Use ratio test because it is more easier: $\lim_{n\to\infty}\frac{a^{n+1}}{{(n+1)}!}/\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{n+1}=0<1$. Hence $\lim_{n\to\infty}\frac{a^n}{n!}=0$. user66607 I hope you get it now.

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  • $\begingroup$ ah thanks i get this one :) $\endgroup$ – chikurin Apr 6 '13 at 3:35

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