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 $!$.

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,
$$P = \frac{2^{8/3}}{e} $$
