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Question Number 158465 by HongKing last updated on 04/Nov/21

Answered by qaz last updated on 04/Nov/21

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{\left(\begin{array}{c}2\mathrm{n}\\ \mathrm{n}\end{array}\right)}{4^{\mathrm{n}}(4\mathrm{n}+5)(2\mathrm{n}+2)}$$$$=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}-1/2\\ \mathrm{n}\end{array}\right){(-1)}^{\mathrm{n}}(\frac{1}{\mathrm{n}+1}-\frac{4}{4\mathrm{n}+5})$$$$=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}-1/2\\ \mathrm{n}\end{array}\right){(-1)}^{\mathrm{n}}{\int }_0^1({\mathrm{x}}^{\mathrm{n}}-{4\mathrm{x}}^{4\mathrm{n}+4})\mathrm{dx}$$$$={\int }_0^1\frac{1}{2}{(1-\mathrm{x})}^{-1/2}-{2\mathrm{x}}^4{(1-{\mathrm{x}}^4)}^{-1/2}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^1{(1-\mathrm{x})}^{-1/2}-{\mathrm{x}}^{1/4}{(1-\mathrm{x})}^{-1/2}\mathrm{dx}$$$$=\frac{1}{2}(\mathrm{B}(1,\frac{1}{2})-\mathrm{B}(\frac{5}{4},\frac{1}{2}))$$$$=1-\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{6\sqrt{2\pi }}$$

Commented by HongKing last updated on 04/Nov/21

$$\mathrm{cool}\mathrm{dear}\mathbf{S}\mathrm{er}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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