I need to prove that :
$$\lim_{n \to \infty} \frac{x^n}{n!}=0$$
I tried to prove it using epsilon-delta, but I still haven't managed to prove it.
I think that maybe with sandwich theorem it possible to prove it.
Any help will be appreciated.
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Do you have that $x>0$? – Aaron Maroja Oct 26 '14 at 17:05
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1$e^x = \sum_{n=0}^\infty\frac{x^n}{n!}$ – Petite Etincelle Oct 26 '14 at 17:26
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4Does this answer your question? Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. – Martin Sleziak May 09 '21 at 13:25
2 Answers
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Hint:
$$\lim_{n\to\infty} \frac{x^{n+1}}{(n+1)!}\frac{n!}{x^n} = \lim_{n\to\infty} \frac{x}{n+1} = 0 <1 $$
Aaron Maroja
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If $\lim \frac{a_{n+1}}{a_n} = c < 1$ then $\lim a_n = 0$, with $a_n > 0$. It's a result. In this case we have $a_n = \frac{x^n}{n!}$. – Aaron Maroja Oct 26 '14 at 17:18
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Let $n_0$ be such that $|x|<n_0$. Then $|\lim_{n\geq 1}\frac {x^n}{n!}|=|\frac{x^{n_0-1}}{(n_0-1)!} \lim_{n\geq n_0}\frac {x^{n-n_0+1}}{n_0\cdots n}|\leq |\frac{x^{n_0-1}}{(n_0-1)!} \lim_{n\geq n_0}\left(\frac {x}{n_0}\right)^{n-n_0+1}| =0$
Giulio Bresciani
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