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Question Number 121985 by Dwaipayan Shikari last updated on 13/Nov/20

Σ_(n=1) ^∞ (1/(n (((4n)),((2n)) )))

∑∞n=11n(4n2n)

Commented by Dwaipayan Shikari last updated on 13/Nov/20

Σ_(n=1) ^∞ (((2n!)^2 )/(n(4n!))) =Σ_(n=1) ^∞ ((Γ(2n+1)Γ(2n+1))/(nΓ(2n+1))) =2Σ_(n=1) ^∞ ((nΓ(2n)Γ(2n+1))/(nΓ(4n+1))) =2Σ_(n=1) ^∞ β(2n,2n+1) =2Σ_(n=1) ^∞ ∫_0 ^1 x^(2n−1) (1−x)^(2n) dx =2∫_0 ^1 Σ_(n=1) ^∞ x^(2n−1) (1−x)^(2n) dx S=x(1−x)^2 +x^3 (1−x)^4 +... −Sx^2 (1−x)^2 =−x^3 (1−x)^4 −x^5 (1−x)^6 −... S(1−x^2 +2x^3 −x^4 )=x(1−x)^2 S=((x(1−x)^2 )/(1−x^2 +2x^3 −x^4 )) So =2∫_0 ^1 ((x(1−x)^2 )/(1−x^2 +2x^3 −x^4 ))dx =−2∫_0 ^1 ((−x^3 +2x^2 −x)/(1−x^2 +2x^3 −x^4 ))dx =−2∫_0 ^1 ((−2x^3 +3x^2 −x)/(1−x^2 +2x^3 −x^4 ))+((x^3 −x^2 )/(1−x^2 +2x^3 −x^4 ))dx =−∫_0 ^1 ((−4x^3 +6x^2 −2x)/(1−x^2 +2x^3 −x^4 ))+2∫((x^2 −x^3 )/(1−x^2 +2x^3 −x^4 ))dx =−[log(1−x^2 +2x^3 −x^4 )]_0 ^1 +2∫((x^2 (1−x))/(1−x^2 (1−x)^2 ))dx

∑∞n=1(2n!)2n(4n!)=∑∞n=1Γ(2n+1)Γ(2n+1)nΓ(2n+1)=2∑∞n=1nΓ(2n)Γ(2n+1)nΓ(4n+1)=2∑∞n=1β(2n,2n+1)=2∑∞n=1∫01x2n−1(1−x)2ndx=2∫01∑∞n=1x2n−1(1−x)2ndxS=x(1−x)2+x3(1−x)4+...−Sx2(1−x)2=−x3(1−x)4−x5(1−x)6−...S(1−x2+2x3−x4)=x(1−x)2S=x(1−x)21−x2+2x3−x4So=2∫01x(1−x)21−x2+2x3−x4dx=−2∫01−x3+2x2−x1−x2+2x3−x4dx=−2∫01−2x3+3x2−x1−x2+2x3−x4+x3−x21−x2+2x3−x4dx=−∫01−4x3+6x2−2x1−x2+2x3−x4+2∫x2−x31−x2+2x3−x4dx=−[log(1−x2+2x3−x4)]01+2∫x2(1−x)1−x2(1−x)2dx

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