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

I've tried multiplying the nominator and the denominator by what is lacking for there to be $3n!$ in the nominator, but that landed me in an algebraic mess.
I've also tried to see if we can extract something like $const^n$ from $2\cdot5\cdot8\cdot\cdots\cdot(3n-4)$, and nothing came out except for a very thin sight of $2^{n/2}$ (I think) that I don't know how to use.
I think we are supposed to use Stirling formula somewhere in the exercise.
Thank you.
 A: 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 $1$.
A: 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 $1$.
A: 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 $1$.
A: An application of Bernoulli's Inequality yields
$$
\begin{align}
\frac{2\cdot5\cdot8\cdot\cdots\cdot(3n-4)}{3^nn!}
&=\frac13\prod_{k=2}^n\frac{3k-4}{3k}\\
&=\frac19\prod_{k=3}^n\frac{k-\frac43}{k}\\
&\ge\frac19\prod_{k=3}^n\left(\frac{k-2}{k}\right)^{2/3}\\[3pt]
&=\frac19\left(\frac2{(n-1)n}\right)^{2/3}\tag1
\end{align}
$$
Apply $\lim\limits_{n\to\infty}n^{1/n}=1$ to get the final limit. This follows from the Binomial Theorem:
$$
\begin{align}
\left(1+\sqrt{\frac2n}\right)^n
&\ge1+n\sqrt{\frac2n}+\frac{n(n-1)}2\frac2n\\
&=n+\sqrt{2n}\\[6pt]
&\ge n\tag2
\end{align}
$$
which implies $n^{1/n}\le1+\sqrt{\frac2n}$
