1
$\begingroup$

I’m working on this limit I guess it had to do with the Riemann sums type of exercise so I am trying to compute it: $$\lim_{n\to\infty} \left(\frac{(2n)\,!}{n^n\,n \,!}\right)^{\frac{1}{2n}}$$

Here is what I have got so far:

$$\lim_{n\to\infty} e^{\frac{1}{2n}{\ln\left(\frac{(2n)\,!}{n^n\,n \,!}\right)}} $$ Now I try to reduce the factorial division: $\frac{(2n)\,!}{n \,!}=\frac{2n(2n-1)(2n-2) \cdots(2n-(n-1))(n)(n-1)\cdots2\cdot1}{(n)(n-1)\cdots2\cdot1}=2n(2n-1)(2n-2) \cdots(2n-(n-1))=\prod_{k=1}^{n-1}(2n-k)$

Therefore I can write the logarithm as : $\ln\left(\frac{\prod_{k=1}^{n-1}(2n-k)}{n^n}\right)=\ln \left(\prod_{k=1}^{n-1}(2n-k)\right)-n\ln(n)$

And I also transform the logarithm into a sum like that should looke like this: $\sum_{k=1}^{n-1}\ln(2n-k)$

So, in the end I got this: $$\lim_{n\to\infty} e^{\frac{1}{2n}{\sum_{k=1}^{n-1}\ln(2n-k)-n\ln(n)}}$$

Now what I can't figure out is how to continue, I want to get some kind of $k/n$ inside the sum and also the limits of the sum I got don't convince me... can someone help me out, or point out any mistakes I could have done?? Thanks in advance ;)

$\endgroup$
2
$\begingroup$

Hint:

If $A=\displaystyle\lim_{n\to\infty} \left(\frac{(2n)\,!}{n^n\,n \,!}\right)^{\frac{1}{2n}}$

$\ln A=\displaystyle\lim_{n\to\infty}\dfrac1{2n}\sum_{r=1}^n\ln\dfrac{(2r-1)(2r)}{n r}$

$2A=\displaystyle\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\dfrac{2(2r-1)}n=\ln2+\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\dfrac{2r-1}n$

Now $\displaystyle\sum_{r=1}^n\ln\dfrac{2r-1}n=\sum_{r=1}^{2n}\ln\dfrac rn-\sum_{r=1}^n\ln\dfrac{2r}n$

$\displaystyle\implies\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\dfrac{2r-1}n=\lim_{n\to\infty}\dfrac1n\sum_{r=1}^{2n}\ln\dfrac rn-\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\dfrac{2r}n$

Set $2n=m$ in the first sum and for both the sum, use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

$\endgroup$
1
$\begingroup$

By ratio-root criteria

$$\frac{b_{n+1}}{b_n} \rightarrow L\implies a_n=b_n^{\frac{1}{n}} \rightarrow L$$

since

$$\left(\frac{(2n+2)\,!}{(n+1)^{n+1}\,(n+1) \,!}\frac {n^n\,n \,!}{(2n)\,!}\right)^{\frac{1}{2}}=\left(\frac{(2n+2)(2n+1)}{(n+1)^2}\frac {1}{\left(1+\frac1n\right)^n}\right)^{\frac{1}{2}}\to\frac{2}{\sqrt e}$$

therefore

$$\left(\frac{(2n)\,!}{n^n\,n \,!}\right)^{\frac{1}{2n}}\to\frac{2}{\sqrt e}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.