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Question Number 85857 by M±th+et£s last updated on 25/Mar/20

Answered by mind is power last updated on 26/Mar/20

$$n^2=\frac{1}{4}({(2n-1)}^2+2(2n-1)+1))$$$$\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{{(-1)}^{n-1}{n}^2}{{(2n-1)}^3}=\sum _{n⩾1}\frac{{(-1)}^{n-1}}{4(2n-1)}+\frac{1}{2}\sum _{n⩾1}\frac{{(-1)}^{n-1}}{{(2n-1)}^2}+\frac{1}{4}\sum _{n⩾1}\frac{{(-1)}^{n-1}}{{(2n-1)}^3}$$$$\sum _{n⩾1}\frac{{(-1)}^{n-1}}{(2n-1)}=\sum _{n⩾0}\frac{{(-1)}^n}{(2n+1)}=S_1$$$${\tan }^{-1}\left(x\right)=\mathrm{\Sigma }\frac{{(-1)}^n{x}^{2n+1}}{2n+1}$$$$S_1={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$$$$S_2=\sum _{n⩾1}\frac{{(-1)}^{n-1}}{{(2n-1)}^2}=\sum _{n⩾0}\frac{{(-1)}^n}{{(2n+1)}^2}$$$$arctan\left(x\right)=\mathrm{\Sigma }\frac{{(-1)}^n{x}^{2n+1}}{2n+1}\Rightarrow {\int }_0^t\frac{{\tan }^{-1}\left(x\right)}{x}dx={\int }_0^t\mathrm{\Sigma }\frac{{(-1)}^n{x}^{2n}}{(2n+1)}dx$$$$=\sum _{n⩾0}\frac{{(-1)}^n{t}^{2n+1}}{{(2n+1)}^2}\Rightarrow S_2={\int }_0^1\frac{{\tan }^{-1}\left(x\right)}{x}dxBypart$$$$S_2={\left[\ln \left(x\right){\tan }^{-1}\left(x\right)\right]}_0^1-{\int }_0^1\frac{\ln \left(x\right)}{1+x^2}dx=-{\int }_0^1\frac{\ln \left(x\right)}{x^2+1}dx=G$$$$S_3=\sum _{n⩾0}\frac{{(-1)}^n}{{(2n+1)}^3}={\int }_0^1\frac{1}{t}{\int }_0^t\frac{{\tan }^{-1}\left(x\right)}{x}dx$$$$={\left[ln\left(t\right){\int }_0^t\frac{{\tan }^{-1}\left(x\right)}{x}dx\right]}_0^1-{\int }_0^1\frac{ln\left(t\right){\tan }^{-1}\left(t\right)}{t}dt$$$$=-{\int }_0^1\frac{\ln \left(t\right)}{t}{\tan }^{-1}\left(t\right)dt=-\left[{\tan }^{-1}\left(t\right)\frac{\ln {}^2\left(t\right)}{2}\right]+\frac{1}{2}{\int }_0^1\frac{{ln}^2\left(t\right)}{1+t^2}dt=S_3$$$$ln\left(t\right)=\underset{x\to 1}{\lim }\frac{\partial }{\partial x}.t^{x-1}$$$$\frac{1}{2}.\underset{y\to 1}{\lim }\frac{{\partial }^2}{\partial y^2}{\int }_0^1\frac{t^{y-1}}{1+t^2}dt\underset{t^2=z}{=}\frac{1}{2}\underset{y\to 1}{\lim }\frac{{\partial }^2}{\partial y^2}{\int }_0^1\frac{z^{\frac{y}{2}-1}}{1+z}.\frac{dz}{2}=\frac{1}{4}\underset{y\to 1}{\lim }\frac{{\partial }^2}{\partial y^2}{\int }_0^1\frac{z^{\frac{y}{2}-1}}{1+z}dz$$$$\frac{1}{4}\underset{y\to 1}{\lim }\frac{{\partial }^2}{\partial y^2}.\frac{\pi }{sin\left(\frac{y\pi }{2}\right)}=\underset{x\to 1}{\lim }.\frac{\partial }{\partial x}.\frac{-{\pi }^{2}cos\left(\frac{\pi x}{2}\right)}{8{sin}^2\left(\frac{\pi x}{2}\right)}$$$$=\frac{{\pi }^3}{16}=S_3$$$$\sum _{n⩾1}\frac{{(-1)}^{n-1}{n}^2}{{(2n-1)}^3}=\frac{S_1}{4}+\frac{S_2}{2}+\frac{S_3}{4}=\frac{\pi }{16}+\frac{G}{2}+\frac{{\pi }^3}{64}=\frac{1}{128}(8\pi +64G+2{\pi }^3)$$$$$$$$$$$$$$$$$$$$$$

Commented by M±th+et£s last updated on 26/Mar/20

$$godblessyousir.nicesolution$$

Commented by mind is power last updated on 26/Mar/20

$$thanxSirhopgoodblessevreyone$$

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