Find $\limsup _{n\to\infty} \bigl(\frac{2\cdot5\cdot8\cdot\cdots\cdot(3n-4)}{3^nn!}\bigr)^{1/n}$

Let the given sequence in question be $a_n$ and $b_n=a_n^n$. Then $$\frac {b_{n+1}}{b_n}=\frac{3n-1}{3(n+1)}$$ which tends to $1$. Thus $a_n=\sqrt[n] {b_n} $ also tends to $1$.

Consider first that $$\prod_{i=2}^n (3i-4)=-\frac{3^n \,\Gamma \left(n-\frac{1}{3}\right)}{\Gamma \left(-\frac{1}{3}\right)}$$ then $$\frac{\prod_{i=2}^n (3i-4)} {3^n \, n! }=-\frac{\Gamma \left(n-\frac{1}{3}\right)}{\Gamma \left(-\frac{1}{3}\right) n!}$$ Take logarithms and use Stirling approximations as well as Taylor series to get $$\log\left(\frac{\prod_{i=2}^n (3i-4)} {3^n \, n! } \right)=-\frac{4}{3} \log \left({n}\right)-\log \left(-{\Gamma \left(-\frac{1}{3}\right)}\right)+\frac{2}{9 n}+O\left(\frac{1}{n^2}\right)$$ Divide by $n$ and show that the limit is just $0$. So, for the whole expression, the limit is $1$.

Squeezing can also give the limit quite quickly.

First note that

• $\bigl(\frac{2\cdot5\cdot8\cdot\cdots\cdot(3n-4)}{3^nn!}\bigr)^{1/n} = \frac{1}{3}\left(\prod_{k=1}^{n}\frac{3k-1}{k}\right)^{\frac{1}{n}}\cdot \underbrace{\frac{1}{\sqrt[n]{3n-1}}}_{\stackrel{n\to\infty}{\longrightarrow}1}$

Now you have

$$\frac{1}{3}\left(\prod_{k=1}^{n}\frac{3k-1}{k}\right)^{\frac{1}{n}} < \frac{1}{3}\left(\prod_{k=1}^{n}\frac{3k}{k}\right)^{\frac{1}{n}} = 1$$

$$\frac{1}{3}\left(\prod_{k=1}^{n}\frac{3k-1}{k}\right)^{\frac{1}{n}} > \frac{1}{3}\left(\prod_{k=2}^n\frac{3k-3}{k}\right)^{\frac{1}{n}} = \frac{1}{\sqrt[n]{3}}\left(\prod_{k=2}^n\frac{k-1}{k}\right)^{\frac{1}{n}}= \frac{1}{\sqrt[n]{3n}}\stackrel{n\to \infty}{\longrightarrow}1$$

Both together give a limit of $1$.