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 =(1/4)((2n−1)^2 +2(2n−1)+1)) Σ_(n=1) ^(+∞) (((−1)^(n−1) n^2 )/((2n−1)^3 ))=Σ_(n≥1) (((−1)^(n−1) )/(4(2n−1)))+(1/2)Σ_(n≥1) (((−1)^(n−1) )/((2n−1)^2 ))+(1/4)Σ_(n≥1) (((−1)^(n−1) )/((2n−1)^3 )) Σ_(n≥1) (((−1)^(n−1) )/((2n−1)))=Σ_(n≥0) (((−1)^n )/((2n+1)))=S_1 tan^(−1) (x)=Σ(((−1)^n x^(2n+1) )/(2n+1)) S_1 =tan^(−1) (1)=(π/4) S_2 =Σ_(n≥1) (((−1)^(n−1) )/((2n−1)^2 ))=Σ_(n≥0) (((−1)^n )/((2n+1)^2 )) arctan(x)=Σ(((−1)^n x^(2n+1) )/(2n+1))⇒∫_0 ^t ((tan^(−1) (x))/x)dx=∫_0 ^t Σ(((−1)^n x^(2n) )/((2n+1)))dx =Σ_(n≥0) (((−1)^n t^(2n+1) )/((2n+1)^2 ))⇒S_2 =∫_0 ^1 ((tan^(−1) (x))/x)dx By part S_2 =[ln (x)tan^(−1) (x)]_0 ^1 −∫_0 ^1 ((ln (x))/(1+x^2 ))dx=−∫_0 ^1 ((ln (x))/(x^2 +1))dx=G S_3 =Σ_(n≥0) (((−1)^n )/((2n+1)^3 ))=∫_0 ^1 (1/t)∫_0 ^t ((tan^(−1) (x))/x)dx =[ln(t)∫_0 ^t ((tan^(−1) (x))/x)dx]_0 ^1 −∫_0 ^1 ((ln(t)tan^(−1) (t))/t)dt =−∫_0 ^1 ((ln (t))/t)tan^(−1) (t)dt=−[tan^(−1) (t)((ln^2 (t))/2)]+(1/2)∫_0 ^1 ((ln^2 (t))/(1+t^2 ))dt=S_3 ln(t)=lim_(x→1) (∂/∂x).t^(x−1) (1/2).lim_(y→1) (∂^2 /∂y^2 )∫_0 ^1 ((t^(y−1) )/(1+t^2 ))dt=_(t^2 =z) (1/2)lim_(y→1) (∂^2 /∂y^2 )∫_0 ^1 (z^((y/2)−1) /(1+z)).(dz/2)=(1/4)lim_(y→1) (∂^2 /∂y^2 )∫_0 ^1 (z^((y/2)−1) /(1+z))dz (1/4)lim_(y→1) (∂^2 /∂y^2 ).(π/(sin(((yπ)/2))))=lim_(x→1) .(∂/∂x).((−π^2 cos(((πx)/2)))/(8sin^2 (((πx)/2)))) =(π^3 /(16))=S_3 Σ_(n≥1) (((−1)^(n−1) n^2 )/((2n−1)^3 ))=(S_1 /4)+(S_2 /2)+(S_3 /4)=(π/(16))+(G/2)+(π^3 /(64))=(1/(128))(8π+64G+2π^3 )](Q85910.png)
$$\left.{n}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)+\mathrm{1}\right)\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{4}\left(\mathrm{2}{n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}={S}_{\mathrm{1}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left({x}\right)=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$${S}_{\mathrm{1}} =\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$$${S}_{\mathrm{2}} =\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${arctan}\left({x}\right)=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\Rightarrow\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}{dx}=\int_{\mathrm{0}} ^{{t}} \Sigma\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}{dx} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\Rightarrow{S}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}{dx}\:{By}\:{part} \\ $$$${S}_{\mathrm{2}} =\left[\mathrm{ln}\:\left({x}\right)\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\:\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\:\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}={G} \\ $$$${S}_{\mathrm{3}} =\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}}\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}{dx} \\ $$$$=\left[{ln}\left({t}\right)\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}{dx}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)\mathrm{tan}^{−\mathrm{1}} \left({t}\right)}{{t}}{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\:\left({t}\right)}{{t}}\mathrm{tan}^{−\mathrm{1}} \left({t}\right){dt}=−\left[\mathrm{tan}^{−\mathrm{1}} \left({t}\right)\frac{\mathrm{ln}\:^{\mathrm{2}} \left({t}\right)}{\mathrm{2}}\right]+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}={S}_{\mathrm{3}} \\ $$$${ln}\left({t}\right)=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\partial}{\partial{x}}.{t}^{{x}−\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}.\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\partial^{\mathrm{2}} }{\partial{y}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{y}−\mathrm{1}} \:}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\underset{{t}^{\mathrm{2}} ={z}} {=}\frac{\mathrm{1}}{\mathrm{2}}\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\partial^{\mathrm{2}} }{\partial{y}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{z}^{\frac{{y}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{z}}.\frac{{dz}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\partial^{\mathrm{2}} }{\partial{y}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{z}^{\frac{{y}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{z}}{dz} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\partial^{\mathrm{2}} }{\partial{y}^{\mathrm{2}} }.\frac{\pi}{{sin}\left(\frac{{y}\pi}{\mathrm{2}}\right)}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}.\frac{\partial}{\partial{x}}.\frac{−\pi^{\mathrm{2}} {cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{\mathrm{8}{sin}^{\mathrm{2}} \left(\frac{\pi{x}}{\mathrm{2}}\right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{16}}={S}_{\mathrm{3}} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }=\frac{{S}_{\mathrm{1}} }{\mathrm{4}}+\frac{{S}_{\mathrm{2}} }{\mathrm{2}}+\frac{{S}_{\mathrm{3}} }{\mathrm{4}}=\frac{\pi}{\mathrm{16}}+\frac{{G}}{\mathrm{2}}+\frac{\pi^{\mathrm{3}} }{\mathrm{64}}=\frac{\mathrm{1}}{\mathrm{128}}\left(\mathrm{8}\pi+\mathrm{64}{G}+\mathrm{2}\pi^{\mathrm{3}} \right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 26/Mar/20
$${god}\:{bless}\:{you}\:{sir}\:.\:{nice}\:{solution}\: \\ $$
Commented by mind is power last updated on 26/Mar/20
$${thanx}\:{Sir}\:{hop}\:{good}\:{bless}\:{evrey}\:{one}\: \\ $$