# Show that $\lim \limits_{n\rightarrow\infty}\frac{n!}{(2n)!}=0$

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 ?

• It looks correct. Apr 6, 2013 at 10:55
• You did it alright Apr 6, 2013 at 10:55
• thanks but maybe i need to show that it is bounded from bellow and above. Apr 6, 2013 at 10:55
• +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. Apr 6, 2013 at 10:58

Hint

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

• this is comment, not an answer? Apr 6, 2013 at 11:00
• I see now, using the Sandwich Theorem can show that the $lim$ is $0$. Thanks and +1 Apr 6, 2013 at 11:06
• "Sandwich Theorem" Jun 3, 2016 at 18:34

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.

• 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. Apr 6, 2013 at 11:05

Another hint based on using series may be that, if the series $$\sum_0^{\infty}u_n$$ is convergent so $u_n\to 0$.

Hint:

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

• nice this is interesting Apr 6, 2013 at 11:25

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}

• Nice this is also a nice solution Apr 6, 2013 at 11:37