Calculate $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\prod_{k=1}^{n}\left(n+3k-1\right)\right)^{\frac{1}{n}}$ I'm need of some assistance regarding a homework question:
$$
\mbox{"Calculate the following:}\quad \lim_{n \to \infty}
\frac{1}{n}\left[%
\prod_{k = 1}^{n}\left(n + 3k -1\right)\right]^{1/n}\
\mbox{"}
$$
Alright so since this question is in the chapter for definite integrals ( and because it is similar to other questions I have answered ) I assumed that I should play a little with the expression inside the limit and change the product to some Riemann sum of a known function.
$0$K so I've tried that but with no major breakthroughs$\ldots$
Any hints and help is appreciated, thanks $!$.
 A: The product $P_n$ may be expressed as follows:
$$P_n = \left  [ \prod_{k=1}^n \left (1+\frac{3 k-1}{n}\right ) \right ]^{1/n} $$
so that
$$\log{P_n} = \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k-1}{n}\right )}$$
as $n \to \infty$, $P_n \to P$ and we have
$$\log{P} = \lim_{n \to \infty} \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k-1}{n}\right )} =  \lim_{n \to \infty} \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k}{n}\right )}$$
which is a Riemann sum for the integral
$$\log{P} = \int_0^1 dx \, \log{(1+3 x)}  = \frac13 \int_1^4 du \, \log{u} = \frac13 [u \log{u}-u]_1^4 = \frac{8}{3} \log{2}-1$$
Therefore,
$$P = \frac{2^{8/3}}{e} $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}\braces{{1 \over n}
\bracks{\prod_{k = 1}^{n}\pars{n + 3k - 1}}^{1/n}\,}} =
\lim_{n \to \infty}\braces{{3 \over n}
\bracks{\prod_{k = 1}^{n}\pars{k + {n - 1 \over 3}}}^{1/n}\,}
:\ {\Large ?}}$

\begin{align}
&\bbox[5px,#ffd]{\prod_{k = 1}^{n}\pars{k + {n - 1 \over 3}}} =
\pars{n + 2 \over 3}^{\overline{n}} =
{{\Gamma\pars{\bracks{n + 2}/3 + n} \over
\Gamma\pars{\bracks{n + 2}/3}}}
\\[5mm] = &\
\left.{{\Gamma\pars{m + n + 1} \over \Gamma\pars{m + 1}}}
\,\right\vert_{\,m\ = \pars{n - 1}/3}\ =\
{\pars{m + n}! \over m!}
\\[5mm]
\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&
{\root{2\pi}\pars{m + n}^{m + n + 1/2}\,\,\expo{-m - n}
\over
\root{2\pi}m^{m + 1/2}\,\,\expo{-m}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &
{n^{m + n + 1/2}\,\,\,\,\pars{1 + 1/3}^{4n/3}
\over
3^{-m - 1/2}\,\,\,n^{m + 1/2}\,\,
\pars{1 - 1/n}^{n/3}}\,\expo{-n}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,\,\, &
3^{n/3}\,n^{n}\,\pars{4 \over 3}^{4n/3}
\expo{1/3}\expo{-n} =
\bracks{3^{1/3}\,n\,\pars{4 \over 3}^{4/3}
\expo{1/\pars{3n}}\,\expo{-1}}^{n}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,\,\, &
\pars{{2^{8/3} \over 3\expo{}}n}^{n}
\end{align}
Finally,
$$
\bbox[5px,#ffd]{\lim_{n \to \infty}\braces{{1 \over n}
\bracks{\prod_{k = 1}^{n}\pars{n + 3k - 1}}^{1/n}\,}} =
\bbx{2^{8/3} \over \expo{}} \approx 2.3359 \\
$$
