I am struggling to find the following limit: $$\lim_{n\to \infty} \frac{1\cdot4\cdot7\cdot...\cdot(3n+1)}{2\cdot5\cdot8\cdot...\cdot(3n+2)}$$
So far I tried to use squeeze theorem and logarithm to proof that the limit is actually 1 but didn't manage to do this.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{n\to \infty}{1 \cdot 4 \cdot 7 \cdot\cdots\cdot\pars{3n + 1} \over 2 \cdot 5 \cdot 8 \cdot\cdots\cdot \pars{3n + 2}}} = \lim_{n\to \infty}{\prod_{i = 0}^{n}\pars{3i + 1} \over \prod_{j = 0}^{n}\pars{3j + 2}} = \lim_{n\to \infty}{\prod_{i = 0}^{n}\pars{i + 1/3} \over \prod_{j = 0}^{n}\pars{j + 2/3}} \\[5mm] = &\ \lim_{n\to \infty}{\pars{1/3}^{\overline{n + 1}} \over \pars{2/3}^{\overline{n + 1}}} = \lim_{n\to \infty}{\Gamma\pars{1/3 + n + 1}/\Gamma\pars{1/3} \over \Gamma\pars{2/3 + n + 1}/\Gamma\pars{2/3}} = {\Gamma\pars{2/3} \over \Gamma\pars{1/3}} \lim_{n\to \infty}{\Gamma\pars{n + 4/3} \over \Gamma\pars{n + 5/3}} \\[5mm] = &\ {\Gamma\pars{2/3} \over \Gamma\pars{1/3}} \lim_{n\to \infty}{\pars{n + 1/3}! \over \pars{n + 2/3}!} = {\Gamma\pars{2/3} \over \Gamma\pars{1/3}} \lim_{n\to \infty}{\root{2\pi}\pars{n + 1/3}^{n + 5/6}\expo{-n - 1/3} \over \root{2\pi}\pars{n + 2/3}^{n + 7/6}\expo{-n - 2/3}} \\[5mm] = &\ {\Gamma\pars{2/3} \over \Gamma\pars{1/3}}\, \lim_{n\to \infty}{n^{n + 5/6}\, \bracks{1 + \pars{1/3}/n}^{n}\expo{-1/3} \over n^{n + 7/6}\,\bracks{1 + \pars{2/3}/n}^{n}\expo{-2/3}} = {\Gamma\pars{2/3} \over \Gamma\pars{1/3}}\, \lim_{n\to \infty}{1 \over n^{1/3}} = \bbx{0} \end{align}