# Math Help - Limit:

1. ## Limit:

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

2. Originally Posted by great_math
$\lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}$
We can write,
$\left( \frac{(2n)!}{(n!)^2}\cdot \frac{n!}{n^n} \right)^{1/n} = \left[ {{2n}\choose n} \right]^{1/n} \cdot \tfrac{1}{n}(n!)^{1/n}$

It is a known-fact that $\tfrac{1}{n}(n!)^{1/n} \to \tfrac{1}{e}$

Also it is well-known that $\sum_{n=0}^{\infty} {{2n}\choose n} x^n = \frac{1}{\sqrt{1-4x}}$ for $|x| < \tfrac{1}{4}$
Now the root test applied to this series must therefore give $\left[ {{2n}\choose n}\right]^{1/n} \to 4$

Thus, the limit should (hopefully be if no mistakes were done) $\frac{4}{e}$.

3. Originally Posted by great_math
$\lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}$
Or

$\lim_{x\to\infty}f(x)^{\frac{1}{x}}=\lim_{x\to\inf ty}\frac{f(x+1)}{f(x)}$

So this limit is equivalent to

$\lim_{n\to\infty}\frac{\frac{(2n+2)(2n+1)(2n)!}{(n +1)^{n+2}n!}}{\frac{(2n)!}{n!n^n}}=\lim_{n\to\inft y}\frac{(2n+2)(2n+1)n^n}{(n+1)^{n+2}}=\lim_{n\to\i nfty}\frac{4n^{n+2}+6n^{n+1}+2n^n}{(n+1)^{n+2}}$

It should be pretty obvious that the second and third terms in the numerator go to zero and as for the other one...it is equivalent ot

$4\lim_{n\to\infty}\frac{1}{\frac{(n+1)^{n+2}}{n^{n +2}}}=\frac{4}{\lim_{n\to\infty}\left(1+\frac{1}{n }\right)^{n+2}}=$ $\frac{4}{\lim_{n\to\infty}\left(1+\frac{1}{n}\righ t)\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right) ^2}=\frac{4}{e\cdot{1}}=\frac{4}{e}$

Or

$n!\sim\sqrt{2\pi{n}}n^ne^{-n}\wedge{(2n)!\sim\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2}$

$\therefore\quad\lim_{n\to\infty}\left(\frac{(2n)!} {n!n^n}\right)^{\frac{1}{n}}\sim\lim_{n\to\infty}\ left(\frac{\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2}{n^n\cdot{\sqrt{2\pi{n}}n^ne^{-n}}}\right)^{\frac{1}{n}}$ $=\lim_{n\to\infty}\frac{\left(4\pi{n}\right)^{\fra c{1}{2n}}4n^2e^{-2}}{(2\pi{n})^{\frac{1}{2n}}n^2e^{-1}}=\frac{4}{e}\sqrt{\lim_{n\to\infty}2^{\frac{1}{ n}}}=\frac{4}{e}$

4. Originally Posted by great_math
$\lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}$
You can also use the fact that $n!\sim_n \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$, well-known as Stirling's approximation.
This gives you $\frac{(2n)!}{n! n^n}\sim_n \frac{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n\sqrt{2\pi n}n^n}=\left(\frac{4}{e}\right)^n\sqrt{2}$, hence, using the fact that it is possible to take the logarithm of an asymptotic equivalence (this is specific to the logarithm):
$\frac{1}{n}\ln\left(\frac{(2n)!}{n! n^n}\right)\sim_n \ln\left(\frac{4}{e}\right)+\frac{1}{n}\ln \sqrt{2} \to_n \ln\left(\frac{4}{e}\right)$.
Using the continuity of the exponential, we deduce:
$\left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\exp\left( \frac{1}{n}\ln\left(\frac{(2n)!}{n! n^n}\right) \right)\to_n \exp\left(\ln\frac{4}{e}\right)$, hence $\lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\frac{4}{e}$.

5. Originally Posted by Laurent
You can also use the fact that $n!\sim_n \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$, well-known as Stirling's approximation.
This gives you $\frac{(2n)!}{n! n^n}\sim_n \frac{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n\sqrt{2\pi n}n^n}=\left(\frac{4}{e}\right)^n\sqrt{2}$, hence, using the fact that it is possible to take the logarithm of an asymptotic equivalence (this is specific to the logarithm):
$\frac{1}{n}\ln\left(\frac{(2n)!}{n! n^n}\right)\sim_n \ln\left(\frac{4}{e}\right)+\frac{1}{n}\ln \sqrt{2} \to_n \ln\left(\frac{4}{e}\right)$.
Using the continuity of the exponential, we deduce:
$\left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\exp\left( \frac{1}{n}\ln\left(\frac{(2n)!}{n! n^n}\right) \right)\to_n \exp\left(\ln\frac{4}{e}\right)$, hence $\lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\frac{4}{e}$.
Looks like I just beat editing you =P...sorry...It appears are methodology for the actual limit is slightly different...so no harm =D

6. Originally Posted by Mathstud28
$\therefore\quad\lim_{n\to\infty}\left(\frac{(2n)!} {n!n^n}\right)^{\frac{1}{n}}\sim\lim_{n\to\infty}\ left(\frac{\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2}{n^n\cdot{\sqrt{2\pi{n}}n^ne^{-n}}}\right)^{\frac{1}{n}}$ $=\lim_{n\to\infty}\frac{\left(4\pi{n}\right)^{\fra c{1}{2n}}4n^2e^{-2}}{(2\pi{n})^{\frac{1}{2n}}n^2e^{-1}}=\frac{4}{e}\sqrt{\lim_{n\to\infty}2^{\frac{1}{ n}}}=\frac{4}{e}$
Be careful with this argument: what about $1+\frac{1}{n}\sim 1$ hence $\left(1+\frac{1}{n}\right)^{1/n}\sim(1)^{1/n}=1$? As I wrote, taking the logarithm of an asymptotic equivalence is correct, while taking the exponential is not (except if the argument of the exponential has a limit, like here, in which case the continuity of the exponential function makes it correct).