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nice-mathematics-prove-that-n-1-2n-n-2-2n-1-2-4n-1-2-pi-m-n-july-1970-




Question Number 114735 by mnjuly1970 last updated on 20/Sep/20
   .... nice  mathematics...      prove  that::                             Σ_(n=1) ^∞ (([ (((2n)),(n) )]^2 )/((2n−1)2^(4n) )) =1−(2/π)                     ... m.n.july. 1970#
.nicemathematicsprovethat::n=1[(2nn)]2(2n1)24n=12πYou can't use 'macro parameter character #' in math mode
Answered by maths mind last updated on 20/Sep/20
Σ_(n≥1) ^2 (([ (((2n)),(n) )]^2 )/((2n−1)2^(4n) ))=Σ(((((2n!)/(n!.n!)))^2 )/((2n−1)2^(4n) ))=S  =1+Σ_(n≥2) (((((2n!)/(n!.n!)))^2 )/((2n−1)2^(4n) ))∣  2n!=2^n .Π_(k=0) ^(n−1) (2k+1)=2^(2n) n!.Π_(k=0) ^(n−1) (k+(1/2))  Σ_(n≥1) ^∞ (([ (((2n)),(n) )]^2 )/((2n−1)2^(4n) ))=Σ_(n≥1) (((((2n!)/(n!.n!)))^2 )/((2n−1)2^(4n) ))=Σ_(n≥1) (((2^(4n) n!^2 Π_(k=0) ^(n−1) (k+(1/2))^2 )/((n!)^4 ))/((2n−1)2^(4n) ))  Σ_(n≥1) ((Π_(k=0) ^(n−1) (k+(1/2))^2 )/((2n−1)(n!)^2 )).1=(1/4)+Σ_(n≥2) ((Π_(k=0) ^(n−1) (k+(1/2))^2 )/((2n−1)n!^2 )).1  =(1/4)+Σ_(n≥2) ((Π_(k=0) ^(n−1) (k+(1/2))Π_(k=0) ^(n−2) (k+(1/2))(n−1+(1/2)))/((2n−1)n!)).(((1)^n )/(n!))  =(1/4)+Σ_(n≥2) ((Π_(k=0) ^(n−1) ((1/2)+k)Π_(k=0) ^(n−2) (k+(1/2))(2n−1))/(2(2n−1)n!)).(1^n /(n!))  Π_(k=0) ^(n−2) (k+(1/2))=−2Π_(k=0) ^(n−1) (k−(1/2))=−2(−(1/2))_n   n!=Π_(k=0) ^(n−1) (1+k)=(1)_n   so We get  S=(1/4)+Σ_(n≥2) ((((1/2))_n .−2(−(1/2))_n )/(2(1)_n )).(1^n /(n!))  recall that _2 F_1 (a,b;c;x)=1+Σ_(n≥1) ((a_n b_n )/c_n ) (x^n /(n!))  ⇒S=(1/4)−Σ_(n≥2) ((((1/2))_n (−(1/2))_n )/((1)_n ))=(1/4)−(   _2 F_1 ((1/2),−(1/2);1;1)−1−((((1/2))_1 (−(1/2))_1 )/((1)_1 )).(1/(1!)))  =(1/4)−(_2 F_1 ((1/2),−(1/2);1;1)−1+(1/4))=1−_2 F_1 ((1/2);−(1/2);1;1)  we use gausse Hypergeometrique theorem    _2 F_1 (a,b;c,1)=((Γ(c)Γ(c−b−a))/(Γ(c−a)Γ(c−b))) this is just result  of β(b,c−b)_2 F_1 (a,b;c;z)=∫_0 ^1 x^(b−1) (1−x)^(c−b+1) (1−zx)^(−a) dx  so we get  2F1((1/2),−(1/2);1;1)=((Γ(1)Γ(1+(1/2)−(1/2)))/(Γ((1/2))Γ((3/2))))=((Γ(1)Γ(1))/((1/2)Γ((1/2))^2 ))=(2/π)  S=1−_2 F_1 ((1/2),−(1/2);1;1)=1−(2/π)
2n1[(2nn)]2(2n1)24n=Σ(2n!n!.n!)2(2n1)24n=S=1+n2(2n!n!.n!)2(2n1)24n2n!=2n.n1k=0(2k+1)=22nn!.n1k=0(k+12)n1[(2nn)]2(2n1)24n=n1(2n!n!.n!)2(2n1)24n=n124nn!2n1k=0(k+12)2(n!)4(2n1)24nn1n1k=0(k+12)2(2n1)(n!)2.1=14+n2n1k=0(k+12)2(2n1)n!2.1=14+n2n1k=0(k+12)n2k=0(k+12)(n1+12)(2n1)n!.(1)nn!=14+n2n1k=0(12+k)n2k=0(k+12)(2n1)2(2n1)n!.1nn!n2k=0(k+12)=2n1k=0(k12)=2(12)nn!=n1k=0(1+k)=(1)nsoWegetS=14+n2(12)n.2(12)n2(1)n.1nn!recallthat2F1(a,b;c;x)=1+n1anbncnxnn!S=14n2(12)n(12)n(1)n=14(2F1(12,12;1;1)1(12)1(12)1(1)1.11!)=14(2F1(12,12;1;1)1+14)=12F1(12;12;1;1)weusegausseHypergeometriquetheorem2F1(a,b;c,1)=Γ(c)Γ(cba)Γ(ca)Γ(cb)thisisjustresultofβ(b,cb)2F1(a,b;c;z)=01xb1(1x)cb+1(1zx)adxsoweget2F1(12,12;1;1)=Γ(1)Γ(1+1212)Γ(12)Γ(32)=Γ(1)Γ(1)12Γ(12)2=2πS=12F1(12,12;1;1)=12π
Commented by mnjuly1970 last updated on 21/Sep/20
very nice thank you sir..
verynicethankyousir..
Commented by maths mind last updated on 21/Sep/20
withe pleasur sir
withepleasursir

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