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Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.

I'm not sure how to go about solving this. Right now I'm trying to use the squeeze theorem. Notice $\frac{1}{n^n} <\frac{2^n}{n!} < \frac{2^n}{3^n}$. If I can prove the limit of the lower and upper sequences $= 0$, then by the squeeze theorem the limit of $\frac{2^n}{n}= 0$. However I dont know how to prove the limit of $\frac{1}{n^n} = 0$ so I'm stuck, perhaps there is a better way to solve this problem than the squeeze theorem?

Thanks.

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marked as duplicate by Martin Sleziak, Pedro Tamaroff, Qiaochu Yuan Oct 23 '12 at 18:57

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.

    
Power over Factorial at ProofWiki –  Martin Sleziak Oct 23 '12 at 18:35
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5 Answers 5

Hint: You can prove by induction that $\dfrac{2^n}{n!}\lt \dfrac{4}{n}$ for every positive integer $n$. Check by hand the first few cases. The induction should proceed nicely after that.

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You need to show $n^n\rightarrow\infty$. Pick any C>1. Then for all $n>C$, one has $n^n> C$, so $\lim_{n\rightarrow\infty}n^n=\infty$

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You have that $\lim_{n\to\infty}{n^{n}}=\infty$, as $$\frac{d}{dx}\left(x^{x}\right)=x^{x}(1+\ln{x})\gt0,\quad \forall x\gt e^{-1}$$

Therefore, as $\lim_{n\to\infty}{n^{n}}=\infty$, we have $\lim_{n\to\infty}{\frac{1}{n^{n}}}=0$ and $\lim_{n\to\infty}{\frac{2^{n}}{3^{n}}}=0$, so your limit evaluates to $0$ by the squeeze theorem.

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You don't need to. You can squeeze like this: $$ 0\leq\frac{2^n}{n!} < \frac{2^n}{3^n}. $$

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Note that each $a_n>0$, and that $$\frac{a_{n+1}}{a_n}=\cfrac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}=\frac{2^{n+1}n!}{2^n(n+1)!}=\frac2{n+1}$$ for all $n$, so $$a_{n+1}=\frac2{n+1}a_n\tag{#}$$ for all $n$. Now, $a_0=\frac{2^0}{0!}=\frac11=1$, so $a_1=1$ by $(\#)$. From this point on, we see from $(\#)$ that each subsequent term will be a multiple of the previous term by a fraction of at most $\frac23$. Noting that $a_0=1<\frac32=\frac32\cdot\frac{2^0}{3^0}$ and that $a_1=1=\frac32\cdot\frac{2^1}{3^1}$, it then follows inductively (by the "From this point on..." statement) that $a_n\leq\frac32\cdot\frac{2^n}{3^n}$ for all $n\geq 2$. Thus, for all $n$, we have $$0<a_n\leq\frac32\cdot\frac{2^n}{3^n}=\frac{2^{n-1}}{3^{n-1}}=\left(\frac23\right)^{n-1},$$ and since $$\lim_{n\to\infty}\left(\frac23\right)^{n-1}=0,$$ then by Squeeze Theorem, it follows that $a_n\to 0$, as well.

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