$\lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})$

Question

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.

Answer 1

We have that

$$\left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)=$$ $$=\frac{1\cdot \color{green}4}{\color{red}2\cdot 3}\,\frac{\color{red}2\cdot \color{green}5}{\color{red}3\cdot \color{green}4}\,\frac{\color{red}3\cdot \color{green}6}{\color{red}4\cdot \color{green}5}\ldots\frac{\color{red}{(n-1)}\cdot \color{green}{(n+2)}}{\color{red}n\cdot \color{green}{(n+1)}}\,\frac{\color{red}n\cdot (n+3)}{(n+1)\cdot \color{green}{(n+2)}}=\frac{n+3}{3(n+1)}$$

Answer 2

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*}

Answer 3

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} $