Simple Series!

$\text{Let the sum of the series be } S$ $1+\dfrac { 1\times 3 }{ 6 } +\dfrac { 1\times 3\times 5 }{ 6\times 8 } +\dfrac { 1\times 3\times 5\times 7 }{ 6\times 8\times 10 } +\dfrac { 1\times 3\times 5\times 7\times 9 }{ 6\times 8\times 10\times 12 }+ \cdots =S$ $\Rightarrow S= 4\times 2 \left(\dfrac{1}{2\times4}+\dfrac { 1\times 3 }{ 2\times 4\times 6 } +\dfrac { 1\times 3\times 5 }{ 2\times 4\times 6\times 8 } +\dfrac { 1\times 3\times 5\times 7 }{2\times 4\times 6\times 8\times 10 } +\dfrac { 1\times 3\times 5\times 7\times 9 }{ 2\times 4\times 6\times 8\times 10\times 12 }+ \cdots\right) \\ \Rightarrow \boxed{S= 4\times2 \left( P \right)} \quad \cdots(1)$ $\text{; where }P=\dfrac { 1 }{ 2\times 4 } +\dfrac { 1\times 3\times 5 }{ 2\times 4\times 6\times 8 } +\dfrac { 1\times 3\times 5\times 7 }{2\times 4\times 6\times 8\times 10 } +\dfrac { 1\times 3\times 5\times 7\times 9 }{ 2\times 4\times 6\times 8\times 10\times 12 }+ \cdots$

$\text{To evaluate } P \text{ we have:}$ $(1-x)^{1/2}=1-\dfrac{x}{2}-\dfrac{x^{2}}{2\times 4}-\dfrac{1\times 3\times x^{3}}{2\times 4\times 6} -\dfrac{1\times 3\times 5\times x^4}{2\times 4\times 6\times 8} \cdots$ $\text{Putting } x=1 \text{ gives }$ $0=1-\dfrac{1}{2}-P \Rightarrow \boxed{P=\dfrac{1}{2}}$ $\text{ Putting this value of } P \text{ in equation } (1) \text{ we get:}$ $\boxed{S= 4\times 2 \left(\dfrac12 \right)=4}$