Infinite limit involving factorial

Question

Finding value of $\displaystyle \lim_{n\rightarrow\infty}\frac{(3n)!}{(n!)^3}$

Try: $$\lim_{n\rightarrow \infty}\frac{\bigg(\frac{3n}{e}\bigg)^{3n}\sqrt{6\pi n}}{\bigg(\frac{n^{3n}}{e^{3n}}\bigg)2\pi n\sqrt{2\pi n}}$$

$$\lim_{n\rightarrow\infty}\frac{27^n}{n}\times \frac{\sqrt{3}}{2\pi}\rightarrow \infty.$$

Could some help me how to solve without using Stirling Approximation. thanks

Answer 1

Use the ratio test : $$\frac{u_{n+1}}{u_n}=\frac{(3n+3)!}{\bigl((n+1)!\bigr)^3}\cdot\frac{(n!)^3}{(3n)!\strut}=\frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3}\sim_\infty\frac{27n^3}{n^3}=27.$$

Answer 2

If you just want to show that the limit diverges to infinity, the following estimate should suffice. $$ \frac{(3n)!}{(n!)^3} = \frac{1}{n^2} \cdot \frac{3n}{n} \cdot \frac{3n-1}{n-1} \cdot \frac{3n-2}{n-1} \cdot \frac{3n-3}{n-1} \cdot \frac{3n-4}{n-2} \dots \frac{6}{2} \cdot \frac{5}{1} \cdot \frac{4}{1} \cdot\frac{3}{1} \cdot 2 \cdot 1 \ge \frac{3^{3n-2}\cdot 2}{n^2} $$

Answer 3

Naively, we have \begin{align} \frac{(3n)!}{(n!)^3} =&\ \frac{\overbrace{(n+1)\cdots(n+n)\cdot(2n+1)\cdots (2n+n)}^{2n \text{ terms}}}{1^2\cdot 2^2\cdots (n-1)^2\cdot n^2}\\ \geq&\ \frac{\overbrace{(n+1)\cdots(n+n)\cdot(2n+1)\cdots (2n+n)}^{2n \text{ terms}}}{ n^{2n}}\\ =&\ \prod^n_{j=1}\left(1+\frac{j}{n} \right)\left(2+\frac{j}{n} \right) \geq 2^n \end{align}

Answer 4

Asymptotic Approximation

Since $$ \begin{align} \frac{\,\frac{(3n)!}{n!^3}\,}{\frac{(3n-3)!}{(n-1)!^3}} &=\frac{3(3n-1)(3n-2)}{n^2}\\ &=\color{#C00}{27}\,\frac{\color{#090}{n-\frac13}}{n}\,\frac{\color{#00F}{n-\frac23}}{n}\tag1 \end{align} $$ we get $$ \begin{align} \frac{(3n)!}{n!^3} &=\color{#C00}{27^n}\color{#090}{\frac{\Gamma\!\left(n+\frac23\right)}{\Gamma\!\left(\frac23\right)}}\color{#00F}{\frac{\Gamma\!\left(n+\frac13\right)}{\Gamma\!\left(\frac13\right)}}\frac1{\Gamma(n+1)^2}\tag2\\ &=\frac{27^n}{\Gamma\!\left(\frac23\right)\Gamma\!\left(\frac13\right)}\frac{\Gamma\!\left(n+\frac23\right)}{\Gamma(n+1)}\frac{\Gamma\!\left(n+\frac13\right)}{\Gamma(n+1)}\tag3\\ &\sim\frac{27^n}{\Gamma\!\left(\frac23\right)\Gamma\!\left(\frac13\right)}\,n^{-1/3}n^{-2/3}\tag4\\ &=\frac{27^n\sin(\pi/3)}{\pi}\frac1n\tag5\\[3pt] &=\frac{\sqrt3}{2\pi}\frac{27^n}{n}\tag6 \end{align} $$ Explanation:
$(2)$: take the product of $(1)$
$(3)$: rearrange
$(4)$: apply Gautschi's Inequality
$(5)$: apply Euler's Reflection Formula
$(6)$:evaluate

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A Simpler Proof Of Divergence

By $(1)$, the ratio of the terms of the sequence tends to $27$. This means the sequence diverges.

Answer 5

A factor of $n!$ cancels, leaving

$$\frac{3n(3n-1)\cdots (2n+1)}{n!}\cdot \frac{2n(2n-1)\cdots (n+1)}{n!} \ge \frac{3n(3n-1)\cdots (2n+1)}{n!}\cdot 1 \ge \frac{(2n)^n}{n^n} = 2^n.$$

The limit is $\infty.$