How to evaluate $\lim \limits_{n \to \infty} \dfrac{27^n (n!)^3}{\big(3n+1\big)!}$ ?
Thanks to Wolfram|Alpha, the answer is $\dfrac{2\pi}{3\sqrt{3}}$. But I am not able to simplify the expression so as to able to evaluate the limit. The $\big(3n+1\big)!$ in the denominator confuses me a lot and stops me from progressing.
In general, is it possible to evaluate the "generalized" limit : $$\lim \limits_{n \to \infty } \dfrac{x^{xn }(n!)^x}{\Gamma(xn+2)}$$, for some $x \in \mathbb{R} \text{ and } \Gamma(x)=(x-1)!=\int \limits_{0} ^ {\infty} t^{x-1}.e^{-t}dt$ ?
Any help will be gratefully acknowledged.
Thanks in Advance :-).
Using Stirling's formula:
$$\Gamma(y+1)\sim \sqrt{2\pi y}\left(\frac{y}{e}\right)^y$$
So:
$$\frac{(x^nn!)^x}{\Gamma(xn+2)}\sim \frac{(2\pi n)^{x/2} x^{nx} n^{nx}/e^{nx}}{(xn+1)\sqrt{2\pi xn}(xn)^{xn}/e^{nx}}=\frac{(2\pi n)^{(x-1)/2}}{\sqrt{x}(xn+1)}$$
So when $x=3$, this is $$\frac{2\pi n}{\sqrt{3}(3n+1)}\to \frac{2\pi}{3\sqrt{3}}$$
The general case appears to be:
$$\frac{(x^nn!)^x}{\Gamma\left(xn+\frac{x+1}{2}\right)}\to\frac{1}{\sqrt{x}}\left(\frac{2\pi}{x}\right)^{(x-1)/2}$$
It's definitely true for $x$ an odd integer, but I'm not sure about other $x$.