Prove that $\prod\limits_{k=0}^n\frac{3k+1}{3k+2}\to0$ by elementary means I'm trying to find limit of sequence $s_n$, but with only elementary operations (eg. without integrals), like basic multiplication or logarithms. 
$$x_n = \frac{3n+1}{3n+2} \wedge s_n = \prod_{k=0}^{n} x_k \wedge n\in\mathbb{N}$$
I know that it's zero, but I have problem with prove. Let $g = \lim_{n \to \infty}s_n$. Then $\ln s_n \rightarrow \ln g$. So.
$$
\sum_{k=0}^{n}\ln \frac{3k+1}{3k+2} = \sum_{k=0}^{n}\ln(3k+1) - \ln(3k+2)
$$
I knot, that I can receive $3k+1$ and $3k-1$, but I don't see solution. Is here some clever logarithmic trick? Or other simple solution?
 A: Show that
$$\left(\dfrac{3k+1}{3k+2}\right)^3 < \dfrac{3k+1}{3k+4}$$
(by expanding and collecting terms in the difference).  Now the telescoping product
$$ \prod_{k=1}^N \left(\dfrac{3k+1}{3k+4}\right)^{1/3} = \dfrac{4^{1/3}}{7^{1/3}} \dfrac{7^{1/3}}{10^{1/3}} \ldots \dfrac{(3N+1)^{1/3}}{(3N+4)^{1/3}} = \dfrac{4^{1/3}}{(3N+4)^{1/3}} $$
goes to $0$ as $N \to \infty$. 
A: This is very similar to Robert Israel's approach:
$$ \left(1-\frac{1}{3k+2}\right)^3 = 1-\frac{3}{3k+2}+\frac{3}{(3k+2)^2}-\frac{1}{(3k+2)^3}\leq 1-\frac{1}{k+2}=\frac{k+1}{k+2} $$
hence:
$$ \prod_{k=1}^{N}\frac{3k+1}{3k+2}\leq \sqrt[3]{\frac{2}{N+2}}\to 0. $$
A: Solution (I hope, correct) made by me.
Notice $\ln x_n \to - \infty \Rightarrow x_n \to 0$ and $x>-1 \Rightarrow \ln(1+x) \leq x$.
$$
\ln\left(\prod_{i=1}^{n}x_i\right) = \sum_{i=1}^{n}\ln x_i = \sum_{k=1}^{n}\ln(1 + \frac{-1}{3k+2}) \leq \sum_{i=1}^{n}\frac{-1}{3k+2}
\leq \sum_{i=1}^{n} \frac{-1}{k}
\to-\infty
$$
So from preliminary observations we know that $\prod_{i=1}^{n}x_i \to 0$.
