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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 =(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 )

n2=14((2n1)2+2(2n1)+1))+n=1(1)n1n2(2n1)3=n1(1)n14(2n1)+12n1(1)n1(2n1)2+14n1(1)n1(2n1)3n1(1)n1(2n1)=n0(1)n(2n+1)=S1tan1(x)=Σ(1)nx2n+12n+1S1=tan1(1)=π4S2=n1(1)n1(2n1)2=n0(1)n(2n+1)2arctan(x)=Σ(1)nx2n+12n+10ttan1(x)xdx=0tΣ(1)nx2n(2n+1)dx=n0(1)nt2n+1(2n+1)2S2=01tan1(x)xdxBypartS2=[ln(x)tan1(x)]0101ln(x)1+x2dx=01ln(x)x2+1dx=GS3=n0(1)n(2n+1)3=011t0ttan1(x)xdx=[ln(t)0ttan1(x)xdx]0101ln(t)tan1(t)tdt=01ln(t)ttan1(t)dt=[tan1(t)ln2(t)2]+1201ln2(t)1+t2dt=S3ln(t)=limx1x.tx112.limy12y201ty11+t2dt=t2=z12limy12y201zy211+z.dz2=14limy12y201zy211+zdz14limy12y2.πsin(yπ2)=limx1.x.π2cos(πx2)8sin2(πx2)=π316=S3n1(1)n1n2(2n1)3=S14+S22+S34=π16+G2+π364=1128(8π+64G+2π3)

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

god bless you sir . nice solution

godblessyousir.nicesolution

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

thanx Sir hop good bless evrey one

thanxSirhopgoodblessevreyone

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