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#

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)

Commented bymnjuly1970 last updated on 21/Sep/20

very nice thank you sir..

Commented bymaths mind last updated on 21/Sep/20

withe pleasur sir