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Question Number 129272 by mnjuly1970 last updated on 14/Jan/21

... nice calculus... evsluate:: Ω=Σ_(n=0) ^∞ (−1)^n (((Γ(n+(3/2)))/(2^n Γ( 2n +2))))=???

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:{nice}\:\:{calculus}... \\ $$$$\:\:\:{evsluate}:: \\ $$$$\:\:\:\:\:\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\left(\frac{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}^{{n}} \:\Gamma\left(\:\mathrm{2}{n}\:+\mathrm{2}\right)}\right)=??? \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 14/Jan/21

Σ_(n=0) ^∞ (−1)^n ((Γ(n+(3/2)))/(Γ(2n+2)))((1/2))^n =2Σ_(n=1) ^∞ (−1)^(n+1) ((Γ(n+(1/2)))/(Γ(2n)))((1/2))^n =2Σ_(n=1) ^∞ (((−1)^(n+1) Γ(n+(1/2))Γ((1/2)))/(Γ(n)Γ(n+(1/2))))2^(1−2n) ((1/2))^n =4(√π)Σ_(n=1) ^∞ (((−1)^(n+1) )/((n−1)!))((1/8))^n =((√π)/2)Σ_(n=0) ^∞ (1/(n!))(−(1/8))^n =((√π)/(2(e)^(1/8) )) (Γ(2n)=2^(2n−1) ((Γ(n)Γ(n+(1/2)))/( (√π))))

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}{n}+\mathrm{2}\right)}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} =\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}{n}\right)}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$=\mathrm{4}\sqrt{\pi}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({n}−\mathrm{1}\right)!}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{n}} =\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)^{{n}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt[{\mathrm{8}}]{{e}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\Gamma\left(\mathrm{2}{n}\right)=\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \frac{\Gamma\left({n}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\:\sqrt{\pi}}\right) \\ $$

Commented by mnjuly1970 last updated on 14/Jan/21

very nice mr payan legendre duplication formula...

$$\:\:\:\:\:\:\:\:\:{very}\:{nice}\:\:\:{mr}\:{payan} \\ $$$$\:\:\:\:\:\:{legendre}\:{duplication}\:{formula}... \\ $$

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