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I have to show that $\lim \limits_{n\rightarrow\infty}\frac{n!}{(2n)!}=0$


I am not sure if correct but i did it like this : $(2n)!=(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))\cdot (n!)$ so I have $$\displaystyle \frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}$$ and $$\lim \limits_{n\rightarrow \infty}\frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}=0$$ is this correct ? If not why ?

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  • $\begingroup$ It looks correct. $\endgroup$
    – Eckhard
    Apr 6, 2013 at 10:55
  • $\begingroup$ You did it alright $\endgroup$ Apr 6, 2013 at 10:55
  • $\begingroup$ thanks but maybe i need to show that it is bounded from bellow and above. $\endgroup$
    – Devid
    Apr 6, 2013 at 10:55
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    $\begingroup$ +1 for showing your progress and thoughts about the problem! Too many people just post their question and expect to get an answer to paste into their homework. $\endgroup$ Apr 6, 2013 at 10:58

5 Answers 5

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Hint

$$0\leq\frac{n!}{(2n)!}\leq\frac{1}{n}$$

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  • $\begingroup$ this is comment, not an answer? $\endgroup$ Apr 6, 2013 at 11:00
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    $\begingroup$ I see now, using the Sandwich Theorem can show that the $lim$ is $0$. Thanks and +1 $\endgroup$
    – Devid
    Apr 6, 2013 at 11:06
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    $\begingroup$ "Sandwich Theorem" $\endgroup$ Jun 3, 2016 at 18:34
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It's correct, but I imagine you're expected to show a bit more work to justify your assertion that $$\lim \limits_{n\rightarrow \infty}\frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}=0$$ An easy way to do this is to bound this sequence of fractions with another, simpler one whose limit you know is 0.

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  • $\begingroup$ This was a exam problem. I was thinking about showing that it is bounded. But i did not cause i could not find a supremum. Hopefully i get some points. Thanks for the answer. $\endgroup$
    – Devid
    Apr 6, 2013 at 11:05
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Another hint based on using series may be that, if the series $$\sum_0^{\infty}u_n$$ is convergent so $u_n\to 0$.

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Hint:

$$ 0 \leq \lim_{n\to \infty}\frac{n!}{(2n)!} \leq \lim_{n\to \infty} \frac{n!}{(n!)^2} = \lim_{k \to \infty, k = n!}\frac{k}{k^2} = \lim_{k \to \infty}\frac{1}{k} = 0.$$

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  • $\begingroup$ nice this is interesting $\endgroup$
    – Devid
    Apr 6, 2013 at 11:25
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If so addressing trivial rigorously I suggest using the notation produtory to fatorial use the formula $n!=\prod_{k=1}^{n}$ . \begin{align} 0\leq \frac{n!}{(2n)!} = & \frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=1}^{2n}k\big)} \\ = & \frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=n+1}^{2n}k\big)\big(\prod_{k=1}^{n}k\big)} \\ = & \frac{1}{\big(\prod_{k=n+1}^{2n}k\big)}\frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=1}^{n}k\big)} \\ = & \frac{1}{\big(\prod_{k=n+1}^{k=2n}k\big)} \\ = & \frac{1}{2n\big(\prod_{k=n+1}^{2n-1}k\big)} \\ = & \frac{1}{2n}\frac{1}{\big(\prod_{k=n+1}^{2n-1}k\big)} \\ \leq & \frac{1}{2n} \end{align}

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  • $\begingroup$ Nice this is also a nice solution $\endgroup$
    – Devid
    Apr 6, 2013 at 11:37

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