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Question Number 152345 by mnjuly1970 last updated on 27/Aug/21

$$$$$$prove...$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{}=\frac{1}{3}+\frac{2\pi \sqrt{3}}{27}...◼$$$$$$

Answered by Kamel last updated on 27/Aug/21

$$$$$$prove...$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{}=\frac{1}{3}+\frac{2\pi \sqrt{3}}{27}...◼$$$$=\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{n\mathrm{\Gamma }(n+1)\mathrm{\Gamma }\left(n\right)}{\mathrm{\Gamma }(2n+1)}=\underset{n=1}{\sum ^{+\mathrm{\infty }}}n{\int }_0^1{t}^{n-1}{(1-t)}^{n}dt$$$$={\int }_0^1\frac{d}{dy}{\left[\underset{n=1}{\sum ^{+\mathrm{\infty }}}{y}^n{(1-t)}^n\right]}_{y=t}dt={\int }_0^1\frac{1-t}{{(1-t(1-t))}^2}dt$$$$={\int }_0^1\frac{u}{{(u^2-u+1)}^2}du={\int }_1^{+\mathrm{\infty }}\frac{u}{{(u^2-u+1)}^2}du$$$$S=\frac{1}{2}{\int }_0^{+\mathrm{\infty }}\frac{udu}{{(u^2-u+1)}^2}$$$$=\frac{1}{2}\frac{d}{da}{\left[{\int }_0^{+\mathrm{\infty }}\frac{du}{(u^2-au+1)}\right]}_{a=1}=\frac{1}{2}\frac{d}{da}{\left[\frac{1}{\sqrt{1-\frac{a^2}{4}}}(\frac{\pi }{2}+Arctan\left(\frac{a}{2\sqrt{1-\frac{a^2}{4}}}\right))\right]}_{a=1}$$$$=\frac{1}{2}[\frac{2}{3\sqrt{3}}(\frac{\pi }{2}+Arctan\left(\frac{1}{\sqrt{3}}\right))-\frac{1}{2\sqrt{3}}\left(\frac{\frac{-1}{\sqrt{3}}-\sqrt{3}}{4}\right)]$$$$=\frac{2\pi }{9\sqrt{3}}+\frac{1}{3}=\frac{1}{3}+\frac{2\pi \sqrt{3}}{27}$$

Commented by mnjuly1970 last updated on 28/Aug/21

$$verhnicesirkamel..tashakor$$

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