Show that $\lim_{n\to\infty}\left(\prod_{k=1}^n\frac{2k+1}{3k-2}\right)^{\frac{1}{n}}=\frac{2}{3}$

Question

How can I show $$\lim_{n\to\infty}\left(\prod\limits_{k=1}^n\frac{2k+1}{3k-2}\right)^{\frac{1}{n}}=\frac{2}{3}$$ with the aid of Gamma function？

Answer 1

We have

$$ \left( \prod_{k=1}^{n} \frac{2k+1}{3k-2} \right)^{\frac{1}{n}} = \frac{2}{3} \left( \prod_{k=1}^{n} \left( 1 + \frac{7}{6k-4} \right) \right)^{\frac{1}{n}}. $$

Therefore the claim follows from

$$ 1 \leq \left( \prod_{k=1}^{n} \left( 1 + \frac{7}{6k-4} \right) \right)^{\frac{1}{n}} \leq \exp\left(\frac{1}{n}\sum_{k=1}^{n} \frac{7}{6k-4} \right) \xrightarrow[n\to\infty]{}1. $$

Here we utilized the inequality $1+x \leq e^{x}$.

Answer 2

We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$ (see here).

We have that $$ \frac{a_{n+1}}{a_n}=\frac{\prod_{k=1}^{n+1}\frac{2k+1}{3k-2}}{\prod_{k=1}^n\frac{2k+1}{3k-2}}=\frac{2(n+1)+1}{3(n+1)-2}\to\frac23 $$ as $n\to\infty$. Hence, $$ \lim_{n\to\infty}\biggl(\prod_{k=1}^n\frac{2k+1}{3k-2}\biggr)^{1/n}=\frac23. $$

Answer 3

We want to compute $$ \frac{2}{3}\lim_{n\to +\infty}\sqrt[n]{\frac{2\,\Gamma\left(\tfrac{1}{3}\right)\Gamma\left(n+\tfrac{3}{2}\right)}{\Gamma\left(\tfrac{1}{2}\right)\Gamma\left(n+\tfrac{1}{3}\right)}}$$ where by Gautschi's inequality the argument of $\sqrt[n]{\cdot}$ behaves like $C\cdot n^{7/6}$ for large values of $n$.
The conclusion is straightforward.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n\to\infty}\pars{\prod_{k = 1}^{n}{2k + 1 \over 3k - 2}}^{1/n} = {2 \over 3}:\ {\large ?}}$.

\begin{align} \lim_{n\to\infty}\pars{\prod_{k = 1}^{n}{2k + 1 \over 3k - 2}}^{1/n} & = \lim_{n\to\infty}\pars{{2^{n} \over 3^{n}} \prod_{k = 1}^{n}{k + 1/2 \over k - 2/3}}^{1/n} = {2 \over 3}\lim_{n\to\infty}\bracks{\pars{3/2}^{\large\overline{n}} \over \pars{1/3}^{\large\overline{n}}}^{1/n} \\[5mm] & = {2 \over 3}\lim_{n\to\infty}\bracks{\Gamma\pars{3/2 + n}/\Gamma\pars{3/2} \over \Gamma\pars{1/3 + n}/\Gamma\pars{1/3}}^{1/n} \\[5mm] & = {2 \over 3}\,\lim_{n\to\infty}\braces{\bracks{\Gamma\pars{1/3} \over\Gamma\pars{3/2}}^{1/n}\bracks{\pars{n + 1/2}! \over \pars{n - 2/3}!}^{1/n}} \\[5mm] &= {2 \over 3}\,\lim_{n\to\infty}\bracks{% \root{2\pi}\pars{n + 1/2}^{n + 1}\expo{-\pars{n + 1/2}} \over \root{2\pi}\pars{n - 2/3}^{n - 1/6}\expo{-\pars{n - 2/3}}}^{1/n} \\[5mm] & = {2 \over 3}\,\lim_{n\to\infty}\braces{% n^{n + 1}\bracks{1 + \pars{1/2}/n}^{\,n + 1}\expo{-\pars{n + 1/2}} \over n^{n - 1/6}\bracks{1 - \pars{2/3}/n}^{\,n - 1/6}\expo{-\pars{n - 2/3}}}^{1/n} \\[5mm] & = {2 \over 3}\,\lim_{n\to\infty}\pars{n^{7/6}{% \expo{1/2}\expo{-1/2} \over\expo{-2/3}\expo{2/3}}}^{1/n} = {2 \over 3}\,\underbrace{\lim_{n\to\infty}n^{7/\pars{6n}}}_{\ds{=\ 1}} = \bbx{2 \over 3} \end{align}