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Question Number 119442 by mnjuly1970 last updated on 24/Oct/20

$$...advancedcalculus...$$$$provethat:\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2\left(\begin{array}{c}2n\\ n\end{array}\right)}\stackrel{???}{=}\frac{\zeta \left(2\right)}{3}$$$$solution::\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2\ast \frac{\left(2n\right)!}{{(n!)}^2}}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n!\ast n!}{n^2\ast \left(2n\right)!}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }(n+1)}{n\mathrm{\Gamma }(2n+1)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\beta (n,n+1)}{n}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^1{x}^{n-1}{(1-x)}^n$$$$={\int }_0^1\frac{1}{x}\mathrm{\Sigma }\frac{{(x-x^2)}^n}{n}dx$$$$=-{\int }_0^1\frac{ln(1-x+x^2)}{x}dx$$$$=-{\int }_0^1\frac{ln(1+x^3)-ln(1+x)}{x}dx$$$$={\int }_0^1\frac{ln(1+x)}{x}dx-{\int }_0^1\frac{ln(1+x^3)}{x}dx$$$$=-{li}_2(-1)-{\int }_0^1\frac{\sum _{n=1}\frac{{(-1)}^{n+1}{x}^{3n}}{n}}{x}dx$$$$=\frac{{\pi }^2}{12}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}{\int }_0^1{x}^{3n-1}dx$$$$$$$$=\frac{{\pi }^2}{12}+\frac{1}{3}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2}=\frac{{\pi }^2}{12}-\frac{{\pi }^2}{36}$$$$=\frac{{\pi }^2}{18}=\frac{\zeta \left(2\right)}{3}✓✓$$$$m.n.july.1970..$$$$$$$$$$$$$$$$$$

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