Explanation
Given $$\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}$$
$$\Rightarrow \dfrac{1}{1\times 2\times3 }+\dfrac{1}{2\times 3\times 4}+\dfrac{1}{3\times 4\times 5}+\dfrac{1}{4\times 5\times 6} $$
$$\Rightarrow \dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+\dfrac{1}{120}$$
$$= \dfrac{20+5+2+1}{120}$$
$$= \dfrac{28}{120}$$
$$= \dfrac{7}{30}$$
Given product $$\displaystyle =\left ( 1-\dfrac{1}{n} \right )\left ( 1-\dfrac{1}{n+1} \right )\left ( 1-\dfrac{1}{n+2} \right )\cdot \cdot \cdot \left ( 1-\dfrac{1}{2n} \right )$$ $$\displaystyle =\dfrac{\left ( n-1 \right )}{n}\times \dfrac{n}{\left ( n+1 \right )}\times \dfrac{n+1}{\left ( n+2 \right )}\cdot \cdot \cdot \left ( \dfrac{2n-1}{2n} \right )=\dfrac{\left ( n-1 \right )}{2n}$$
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