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Question Number 123807 by Dwaipayan Shikari last updated on 28/Nov/20

$$\frac{\left(\frac{1}{2}\right)!\left(1\right)!}{\left(\frac{5}{2}\right)!}-\frac{\left(\frac{3}{2}\right)!\left(2\right)!}{\left(\frac{9}{2}\right)!}+\frac{\left(\frac{5}{2}\right)!3!}{\left(\frac{13}{2}\right)!}-\frac{\left(\frac{7}{2}\right)!4!}{\left(\frac{17}{2}\right)!}+....$$

Commented by Dwaipayan Shikari last updated on 28/Nov/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}\frac{\mathrm{\Gamma }(n+\frac{1}{2})\mathrm{\Gamma }(n+1)}{\mathrm{\Gamma }(2n+\frac{3}{2})}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}\beta (n+\frac{1}{2},n+1)$$$$={\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{x}^{n-\frac{1}{2}}{(1-x)}^{n}dx$$$$={\int }_0^1\frac{1}{\sqrt{x}}\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{x}^n{(1-x)}^{n}dx$$$$={\int }_0^1\frac{1}{\sqrt{x}}.\frac{1}{1+x(1-x)}dxx=t^2\Rightarrow 1=2t\frac{dt}{dx}$$$$=2{\int }_0^1\frac{1}{1+t^2-t^4}dx=-2{\int }_0^1\frac{1}{t^4-t^2-1}dt$$$$=-2{\int }_0^1\frac{1}{t^2-{\left(\sqrt{\frac{1+\sqrt{3}}{2}}\right)}^2}+2{\int }_0^1\frac{1}{t^2-{\left(\sqrt{\frac{1-\sqrt{3}}{2}}\right)}^2}$$$$=\frac{2}{\sqrt{\frac{1-\sqrt{3}}{2}}}log\left(\frac{t-\sqrt{\frac{1-\sqrt{3}}{2}}}{t+\sqrt{\frac{1-\sqrt{3}}{2}}}\right)-\frac{2}{\sqrt{\frac{1+\sqrt{3}}{2}}}log\left(\frac{t-\sqrt{\frac{1+\sqrt{3}}{2}}}{t+\sqrt{\frac{1+\sqrt{3}}{2}}}\right)$$$$$$

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