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

      prove...       S =Σ_(n=1) ^∞ (1/) = (1/3) +((2π (√3))/(27)) ...■

prove...S=n=11=13+2π327...

Answered by Kamel last updated on 27/Aug/21

      prove...       S =Σ_(n=1) ^∞ (1/) = (1/3) +((2π (√3))/(27)) ...■        =Σ_(n=1) ^(+∞) ((nΓ(n+1)Γ(n))/(Γ(2n+1)))=Σ_(n=1) ^(+∞) n∫_0 ^1 t^(n−1) (1−t)^n dt     =∫_0 ^1 (d/dy)[Σ_(n=1) ^(+∞) y^n (1−t)^n ]_(y=t) dt=∫_0 ^1 ((1−t)/((1−t(1−t))^2 ))dt     =∫_0 ^1 (u/((u^2 −u+1)^2 ))du=∫_1 ^(+∞) (u/((u^2 −u+1)^2 ))du  S=(1/2)∫_0 ^(+∞) ((udu)/((u^2 −u+1)^2 ))     =(1/2)(d/da)[∫_0 ^(+∞) (du/((u^2 −au+1)))]_(a=1) =(1/2)(d/da)[(1/( (√(1−(a^2 /4)))))((π/2)+Arctan((a/( 2(√(1−(a^2 /4)))))))]_(a=1)      =(1/2)[ (2/(3(√3)))((π/2)+Arctan((1/( (√3)))))−(1/( 2(√3)))(((((−1)/( (√3)))−(√3))/4))]    =((2π)/( 9(√3)))+(1/3)=(1/3)+((2π(√3))/(27))

prove...S=n=11=13+2π327...=+n=1nΓ(n+1)Γ(n)Γ(2n+1)=+n=1n01tn1(1t)ndt=01ddy[+n=1yn(1t)n]y=tdt=011t(1t(1t))2dt=01u(u2u+1)2du=1+u(u2u+1)2duS=120+udu(u2u+1)2=12dda[0+du(u2au+1)]a=1=12dda[11a24(π2+Arctan(a21a24))]a=1=12[233(π2+Arctan(13))123(1334)]=2π93+13=13+2π327

Commented by mnjuly1970 last updated on 28/Aug/21

verh nice sir kamel..tashakor

verhnicesirkamel..tashakor

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