For the second time I've had trouble calculating this limit
$$\lim\limits_{n\rightarrow\infty}\frac{(n＋2)(n＋5)\cdot\ldots\cdot(n+3r-1)\cdot\ldots\cdot(4n－1)}{(n＋1)(n＋4)\cdot\ldots\cdot(n+3r-2)\cdot\ldots\cdot(4n－2)}$$
This limit can be calculated by limitation method but it is not a smart method.
When using the inventory method, you get a result:
$$\lim\limits_{n\rightarrow\infty}\frac{(n＋2)(n＋5)\cdot\ldots\cdot(n+3r-1)\cdot\ldots\cdot(4n－1)}{(n＋1)(n＋4)\cdot\ldots\cdot(n+3r-2)\cdot\ldots\cdot(4n－2)}\quad＝\sqrt[3]{4}$$
I'm looking for a clever way to calculate the limit in a straightforward way
It is the second time that I try to ask a question on this site. I apologize if my question is not accurate.
Best Answer
Using $$ \Gamma(x) \approx \sqrt{2 \pi} \, x^{x - 1/2} \, e^{-x} \, \left(1 + \frac{1}{12 \, x} + \mathcal{O}\left(\frac{1}{x^2}\right) \right) $$ then it can be shown that $$ \frac{\Gamma\left(\frac{n+1}{3}\right)}{\Gamma\left(\frac{n+2}{3}\right)} \approx \left(\frac{n}{3}\right)^{-1/3}. $$
Now, \begin{align} L &= \lim_{n \to \infty} \, \prod_{j=1}^{n} \left\{ \frac{n + 3 j -1}{n + 3 j -2} \right\} \\ &= \lim_{n \to \infty} \frac{\left( \frac{n+2}{3} \right)_{n} }{ \left(\frac{n+1}{3}\right)_{n} } = \lim_{n \to \infty} \frac{\Gamma\left( \frac{n+1}{3} \right) \, \Gamma\left( \frac{4n+2}{3} \right)}{\Gamma\left( \frac{n+2}{3} \right) \, \Gamma\left( \frac{4n+1}{3} \right)} \\ &= \lim_{n \to \infty} \left( \left(\frac{n}{3}\right)^{-1/3} \, \left(\frac{4 n}{3}\right)^{1/3} \right) \\ &= \lim_{n \to \infty} \sqrt[3]{4} \end{align} which gives $$ \lim_{n \to \infty} \, \prod_{j=1}^{n} \left\{ \frac{n + 3 j -1}{n + 3 j -2} \right\} = \sqrt[3]{4}. $$