Limit of a Sequence involving factorials

Question

What is the limit of the sequence given by $x_n=\dfrac{(3n)!}{(n!)^3}$

We have to evaluate $\displaystyle \lim_{n \to \infty} x_n$

I think the limit will tend to infinity because the numerator is $n!(n+1)....(n+n)(2n+1)...(3n)$ product of $3$ numbers each divisible by $n!$ and it keeps on increasing.

Also what will be the limit of $y_n= \sqrt[n]{x_n}$ ?

Answer 1

$$\lim_{n\rightarrow+\infty}\frac{(3n)!}{(n!)^3}=\lim_{n\rightarrow+\infty}\binom{3n}{n,n,n}=+\infty.$$ By the way, always there is the following. $$\frac{(3n)!}{(n!)^3}=\frac{3n}{n}\cdot\frac{3n-1}{n-1}\cdot...\cdot\frac{2n+1}{1}\cdot\frac{2n}{n}\cdot\frac{2n-1}{n-1}\cdot...\cdot\frac{n+1}{1}>n\rightarrow+\infty.$$

Answer 2

$$x_n=\dfrac{(3n)!}{(n!)^3}\implies \log(x_n)=\log((3n)!)-3\log(n!)$$ Now, use Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+O\left(\frac{1}{p}\right)$$ Apply and get $$\log(x_n)=3n \log (3)+\log \left(\frac{\sqrt{3}}{2 \pi n}\right)+O\left(\frac{1}{n}\right)$$ For the fun of it (you can do it with your pocket calculator) $$x_5=756756\implies \log(x_5)\approx 13.5368$$ while the above expansion would give $$x_5\approx 15 \log (3)+\log \left(\frac{\sqrt{3}}{10 \pi }\right)\approx 13.5812$$