Limit involving $n$ th root of expression with factorials: $\lim_{n\to\infty}\frac1n\left\{\frac{(2n)!}{n!}\right\}^{\frac{1}{n}}$

Question

I have a problem that finding the values of limit for the following expression. $$ \lim_{n\to\infty}\dfrac{1}{n}\left\{\dfrac{(2n)!}{n!}\right\}^{\dfrac{1}{n}} $$

Thank you.

Answer 1

Assume that $L$ is the limit. $$n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\quad\Rightarrow\quad (2n)!\sim \sqrt{2\pi\,(2n)}\left(\frac{2n}{e}\right)^{2n}$$ Substitute these in and simplify. We can substitute them in because the error in each approximation becomes $0$ as $n\to+\infty$. $$L=\lim_{n\to+\infty}\frac{1}{n}\frac{\left(\sqrt{4\pi\,n}\left(\frac{2n}{e}\right)^{2n}\right)^{1/n}}{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^{1/n}}$$ $$L=\lim_{n\to+\infty}\frac{1}{n}\frac{(\sqrt{4\pi n})^{1/n}\left(\frac{2n}{e}\right)^2}{(\sqrt{2\pi n})^{1/n}\left(\frac{n}{e}\right)}$$ $$L=\lim_{n\to+\infty}\frac{(\sqrt{2})^{1/n}\cdot4n}{e\cdot n}$$ The $(\sqrt{2})^{1/n}$ should become $1$ and the other two $n$'s should cancel, which gives us the limit. $$\therefore\quad L=\frac{4}{e}\approx1.472$$

http://bit.ly/1s8IUfO

Wolfram Alpha seems to agree with me but this is the first time I've done a limit like this so I'm not sure how much I trust myself. Hope I helped!

Answer 2

Using logarithms and the simplest form of Stirling approximation of $n!$ seems to make things slightly simpler. Since, for large values of $n$, $$\log(n!) \simeq n\log(n)-n$$ then $$\log\frac{2n!}{n!}\simeq\Big(2n\log(2n)-2n\Big)-\Big(n\log(n)-n\Big)=n \log(n)+(2\log(2)-1)n$$ So,$$\frac {1}{n}\log\frac{2n!}{n!}\simeq \log(n)+(2\log(2)-1)$$ $$\left\{\dfrac{(2n)!}{n!}\right\}^{\frac{1}{n}}\simeq n e^{2\log(2)-1}$$ $$\dfrac{1}{n}\left\{\dfrac{(2n)!}{n!}\right\}^{\dfrac{1}{n}}\simeq e^{2\log(2)-1}=\frac{4}{e}$$

You could easily check that, for $n=4$, the value of the expression is less than $10$% larger than the limit.

Answer 3

You can rewrite your limit in the form $$L=\lim_n \left( \frac{(2n)!}{n!n^n} \right)^{\frac{1}{n}}$$ to compute this limit, you can try to compute $$\lim_n \frac{\frac{(2(n+1))!}{(n+1)!(n+1)^{n+1}}}{\frac{(2n)!}{n!n^n}}$$ and use that if this limit exists, then it must be equal to $L$. And now $$\lim_n \frac{(2(n+1))!}{(n+1)!(n+1)^{n+1}} \frac{n!n^n}{(2n)!} = \lim_n \frac{(2n+2)(2n+1)}{(n+1)^2} \left( \frac{n}{n+1} \right)^n = 4e^{-1}$$

So your limit is $\frac{4}{e}$.

Answer 4

By Stirling's approximation, we have $$\sqrt[n]{n!} \sim_{\infty} \frac{n}{e} \,\text{ and }\, \sqrt[n]{n!} \sim_{\infty} \Big(\frac{2n}{e}\Big)^2$$ so $$\frac{1}{n}\sqrt[n]{ \frac{2n!}{n!} } \sim_\infty \frac{1}{n}\big(\frac{2n}{e}\Big)^2 \frac{e}{n} = \frac{4}{e}.$$

Answer 5

Thank you for some answer to this question.

$$ \lim_{n\to\infty}\dfrac{1}{n}\left\{\dfrac{(2n)!}{n!}\right\}^{\dfrac{1}{n}} \Leftrightarrow \lim_{n\to\infty}\left(\dfrac{(2n)!}{n!n^n}\right)^{\dfrac{1}{n}} $$

Then, \begin{eqnarray} \log\left(\dfrac{(2n)!}{n!n^n}\right)^{\dfrac{1}{n}}&=&\dfrac{1}{n}\log\dfrac{(2n)!}{n!n^n}\\ &=&\dfrac{1}{n}\log \left(\dfrac{1}{n^n}(2n)\cdots(n+1)\right)\\ &=&\dfrac{1}{n}\log \left(\left(1+\frac{1}{n}\right)\cdots \left(1+\frac{n}{n}\right)\right)\\ &\longrightarrow_{n\to\infty}&\int_0^1 \log(1+x)dx\\ &=&\log\frac{4}{e} \end{eqnarray}

So, the answer is $$\dfrac{4}{e}.$$