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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 ;)

L. Sandoval
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  • https://math.stackexchange.com/questions/475786/how-to-compute-lim-n-rightarrow-infty-frac1n-left-2n12n2-cdots2nn/475791#475791 https://math.stackexchange.com/questions/465075/find-lim-limits-n-to-infty-frac1n-sum-limits2n-r-1-fracr-sq https://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 – lab bhattacharjee Jun 25 '18 at 16:05
  • @L.Sandoval Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Aug 05 '18 at 20:58

2 Answers2

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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 $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

lab bhattacharjee
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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

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

user
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