$\lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})$ Evaluate  
$\lim_{n \rightarrow  \infty} \left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)$
My attempts : i know  that  $1 -   \frac {2}{k(k+1)} = \frac{(k+2)(k-1)}{k(k+1)}$
After that I'm not able to proceed  further.
Any hints/solution will be apprecaited.
Thanks you and thanks in advance.
 A: Hint: 
\begin{eqnarray*}
\frac{ 1\times 4}{ 3\times 2}\times \frac{ 2\times 5}{ 4\times 3}\times \frac{ 3 \times 6}{ 5 \times 4}\times \cdots 
\end{eqnarray*}
A: A formal proof
with product notation.
$lim_{n \rightarrow  \infty} (  1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})
$
$\begin{array}\\
\prod_{k=2}^n (1-\dfrac{2}{k(k+1)})
&=\prod_{k=2}^n \dfrac{k(k+1)-2}{k(k+1)}\\
&=\prod_{k=2}^n \dfrac{k^2+k-2}{k(k+1)}\\
&=\prod_{k=2}^n \dfrac{(k+2)(k-1)}{k(k+1)}\\
&=\dfrac{\prod_{k=2}^n (k+2)\prod_{k=2}^n (k-1)}{\prod_{k=2}^n k\prod_{k=2}^n (k+1)}\\
&=\dfrac{\prod_{k=4}^{n+2}k\prod_{k=1}^{n-1}k}{\prod_{k=2}^n k\prod_{k=3}^{n+1}k}\\
&=\dfrac{\prod_{k=4}^{n+2}k\prod_{k=1}^{n-1}k}{\prod_{k=3}^{n+1}k\prod_{k=2}^n k}\\
&=\dfrac{(n+2)(1)}{3n}\\
&=\dfrac{n+2}{3n}\\
&=\dfrac13+\dfrac{2}{3n}\\
\end{array}
$
A: We have that
$$\left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)=$$
$$=\frac{1\cdot \color{green}4}{\color{red}2\cdot 3}\,\frac{\color{red}2\cdot \color{green}5}{\color{red}3\cdot \color{green}4}\,\frac{\color{red}3\cdot \color{green}6}{\color{red}4\cdot \color{green}5}\ldots\frac{\color{red}{(n-1)}\cdot \color{green}{(n+2)}}{\color{red}n\cdot \color{green}{(n+1)}}\,\frac{\color{red}n\cdot (n+3)}{(n+1)\cdot \color{green}{(n+2)}}=\frac{n+3}{3(n+1)}$$
