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Question Number 117365 by Dwaipayan Shikari last updated on 11/Oct/20

Express ((1/2))!  in terms of infinite series

$${Express}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!\:\:{in}\:{terms}\:{of}\:{infinite}\:{series} \\ $$

Commented by Dwaipayan Shikari last updated on 11/Oct/20

((1/2))!=((√π)/2)  1−(1/3)+(1/5)−(1/7)+(1/9)−...=(π/4)  (√(1−(1/3)+(1/5)−(1/7)+(1/9)−....))=((√π)/2)

$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{9}}−...=\frac{\pi}{\mathrm{4}} \\ $$$$\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{9}}−....}=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$

Answered by mindispower last updated on 11/Oct/20

((1/2))=(√π)=(√π).Σ_(k≥1) (1/2^k )

$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}=\sqrt{\pi}.\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{{k}} } \\ $$

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