Calculate $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\prod_{k=1}^{n}\left(n+3k-1\right)\right)^{\frac{1}{n}}$

Question

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 $!$.

Answer 1

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,

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

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}\braces{{1 \over n} \bracks{\prod_{k = 1}^{n}\pars{n + 3k - 1}}^{1/n}\,}} = \lim_{n \to \infty}\braces{{3 \over n} \bracks{\prod_{k = 1}^{n}\pars{k + {n - 1 \over 3}}}^{1/n}\,} :\ {\Large ?}}$

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\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, $$ \bbox[5px,#ffd]{\lim_{n \to \infty}\braces{{1 \over n} \bracks{\prod_{k = 1}^{n}\pars{n + 3k - 1}}^{1/n}\,}} = \bbx{2^{8/3} \over \expo{}} \approx 2.3359 \\ $$