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Question Number 189323 by mnjuly1970 last updated on 14/Mar/23

Answered by witcher3 last updated on 14/Mar/23

⇔Σ((Γ(n)Γ(n))/(Γ(2n).2^(−n) ))=Σ_(n≥1) ((β(n,n))/2^(−n) ) =Σ_(n≥1) ∫_0 ^1 ((x^(n−1) (1−x)^(n−1) )/(2^(−n) )) =2∫_0 ^1 Σ_(n≥0) (2x(1−x))^n 2x(1−x)<(1/2) =2∫_0 ^1 (1/(1−2x(1−x))) =2∫_0 ^1 (dx/(2x^2 −2x+1)) =2∫_0 ^1 (dx/(2((x−(1/2))+(1/4))))=4∫_0 ^1 (dx/((2x−1)^2 +1)) =4tan^(−1) (1)=π

$$\Leftrightarrow\Sigma\frac{\Gamma\left(\mathrm{n}\right)\Gamma\left(\mathrm{n}\right)}{\Gamma\left(\mathrm{2n}\right).\mathrm{2}^{−\mathrm{n}} }=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\beta\left(\mathrm{n},\mathrm{n}\right)}{\mathrm{2}^{−\mathrm{n}} } \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}−\mathrm{1}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{2}^{−\mathrm{n}} \:} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\mathrm{2x}\left(\mathrm{1}−\mathrm{x}\right)\right)^{\mathrm{n}} \\ $$$$\mathrm{2x}\left(\mathrm{1}−\mathrm{x}\right)<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−\mathrm{2x}\left(\mathrm{1}−\mathrm{x}\right)} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{2}\left(\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\right)}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\mathrm{4tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\pi \\ $$

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