1
$\begingroup$

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 :-).

  • 5
    https://en.wikipedia.org/wiki/Stirling's_approximation – 2017-02-18
  • 1
    use the Stirling formula for factorial – 2017-02-18
  • 0
    https://fr.wikipedia.org/wiki/Formule_de_Stirling – 2017-02-18
  • 0
    What do you mean "not really know Stirling's formula" ? Either you know it (it's in your knowledge base, or in your available methods to solve exercises), or you don't (don't know it, don't have the right to use it... ). – 2017-02-18
  • 0
    @NicolasFRANCOIS I have heard about it a lot and I know about its use, but I do not know how to implement it as I am not thorough with its concept. In short, I do not know Stirling's Approximation". So I am looking for a simpler way, if there is any, to evaluate the limit. – 2017-02-18
  • 0
    And BTW, $(xn+1)!$ is defined only for integer values of $x$ (in fact, $xn+1$)... At least for people who don't "really know" Stirling :-) – 2017-02-18
  • 1
    @Nirbhay : the fact that $\pi$ is present in the answer tells you there may not be "elementary ways" to solve this problem. Not without reinventing Stirling, anyway :-) – 2017-02-18
  • 0
    @Nirbhay : Ahhh ! So you know $\Gamma$ function, but not Stirling ? Time you complete your knowledge base, by studying how to demonstrate and use this remarkable formula ! – 2017-02-18
  • 0
    @NicolasFRANCOIS I am just 14 year old and a tenth grader ..... – 2017-02-18
  • 1
    It's probably more like: $$\lim \limits_{n \to \infty } \dfrac{x^{xn }(n!)^x}{\Gamma(xn+2)}.$$ That is, $x^{xn}$ rather than $x^{3n}$. – 2017-02-18
  • 0
    @ThomasAndrews Thanks for your suggestion :-) – 2017-02-18

1 Answers 1

3

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 $x$.