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

Prove that $\frac{2^n}{n!}$ converges 0.

I can see why, I just don't get how exactly to do convergence proofs. Right now I have:

For $n>6$, $|\frac{2^n}{n!}-0|=\frac{2^n}{n!}<\frac{2^n}{3^n}$

and

assuming $\frac{2^n}{3^n}<\epsilon$, $n<\frac{\ln\epsilon}{\ln\frac2 3}$

Not sure if the last step is even right...

(This was an exam question today)

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marked as duplicate by Martin Sleziak, Chris Taylor, BenjaLim, Chris Eagle, Asaf Karagila Oct 12 '12 at 9:10

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4 Answers 4

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I'm pretty sure that last one need to be $n > \frac{\ln \varepsilon}{\ln \frac{2}{3}}$. But then that this works. For every $\varepsilon$ you give an explicit way to find $N\left(= \frac{\ln \varepsilon}{\ln \frac{2}{3}})\right)$ such that for all $n > N$ we have $|x_n - 0| < \varepsilon$. Definitions, ta da!

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Where did I go wrong with the $n > \frac{\ln \varepsilon}{\ln \frac{2}{3}}$? –  fhyve Oct 12 '12 at 7:23
  • One has $\lim_{n\to\infty}\frac2n=0$ (surely you can prove this).
  • If $\lim_{n\to\infty}f_n=0$ and $p_n=\prod_{1\leq i\leq n}f_i$ then $\lim_{n\to\infty}p_n=0$. For instance because there exists $N$ with $|f_n|<\frac12$ for $n>N$, and then $|p_n|<\frac C{2^n}$ for $n>N$, where $C=2^N|p_N|$.
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Let $a_{n}=\frac{2^n}{n!}$. It is easy to see that $a_n$ is a positive decreasing sequence hence it has a limit $l$. It follows from the equation $a_n=\frac{2}{n}a_{n-1},n\gt1$ that $l=\lim_{n\to\infty}a_n=(\lim_{n\to\infty}\frac{2}{n})(\lim_{n\to\infty}a_{n-1})=0\times l=0$.

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We first see that $(n-1)! > 2^n$ , for $n=6$ ,

moreover $k≥6$ and $(k-1)!>2^k$ implies $k! > k(2^k) > 2(2^k)=2^(k+1)$ i.e., $((k+1)-1)! > 2^(k+1)$ ; hence $(n-1)! > 2^n$ is true for all $n≥6$ , by principle of mathematical induction ; so that $n!> n (2^n)$ is true for all $n ≥ 6$ i.e., $2^n /n! < 1/n$ is true for all $n ≥6$ and since $n→∞$ implies $n≥6$ and $1/n→0$ and since $2^n / n!$ is never negative , we conclude $2^n / n!$ tends to 0 as $n→∞$

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