How to calculate $ \lim_{n\to\infty}\prod_{i=0}^n (1 - \frac{1}{3i+2})$? 
How to calculate 
  $$
\lim_{n\to\infty}\prod_{i=0}^n (1 - \frac{1}{3i+2})\quad?
$$ 

By transforming it with natural logarithm, the product is equal to 
$$
\large e^{\sum_{i=0}^n\ln(1-\frac{1}{3i+2})},
$$ 
so it boils down to calculating limit of the sum
$$
\sum_{i=0}^n\ln(1-\frac{1}{3i+2}).$$ I have no idea how to proceed.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\lim_{n \to \infty}\prod_{i = 0}^{n}\pars{1 - {1 \over 3i + 2}}} =
\lim_{n \to \infty}{\prod_{i = 0}^{n}\pars{i + 1/3} \over \prod_{i = 0}^{n}\pars{i + 2/3}} =
\lim_{n \to \infty}{\pars{1/3}^{\overline{n + 1}} \over \pars{2/3}^{\overline{n + 1}}}
\\[5mm] = &\
\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}{\pars{n + 1/3}! \over \pars{n + 2/3}!}
\\[5mm] = &\
{\Gamma\pars{2/3}\Gamma\pars{1/3} \over \Gamma^{2}\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] = &\
{\pi/\sin\pars{\pi/3} \over \Gamma^{2}\pars{1/3}}
\lim_{n \to \infty}
{n^{n + 5/6}\bracks{1 + \pars{1/3}/n}^{n + 5/6}\expo{1/3} \over
n^{n + 7/6}\pars{n + 2/3}^{n + 7/6}}
\\[5mm] = &\
{2\root{3}\pi\expo{1/3} \over 3\Gamma^{2}\pars{1/3}}
\lim_{n \to \infty}{1 \over n^{1/3}} = \bbx{0}
\end{align}
