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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#
$$\:\:\:….\:{nice}\:\:{mathematics}… \\ $$$$ \\ $$$$\:\:{prove}\:\:{that}::\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left[\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\right]^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\mathrm{2}^{\mathrm{4}{n}} }\:=\mathrm{1}−\frac{\mathrm{2}}{\pi}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{m}.{n}.{july}.\:\mathrm{1970}# \\ $$
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)
$$\underset{{n}\geqslant\mathrm{1}} {\overset{\mathrm{2}} {\sum}}\frac{\left[\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\right]^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\mathrm{2}^{\mathrm{4}\boldsymbol{{n}}} }=\Sigma\frac{\left(\frac{\mathrm{2}{n}!}{{n}!.{n}!}\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\mathrm{2}^{\mathrm{4}{n}} }={S} \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(\frac{\mathrm{2}{n}!}{{n}!.{n}!}\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\mathrm{2}^{\mathrm{4}{n}} }\mid\:\:\mathrm{2}{n}!=\mathrm{2}^{{n}} .\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{k}+\mathrm{1}\right)=\mathrm{2}^{\mathrm{2}{n}} {n}!.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left[\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\right]^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\mathrm{2}^{\mathrm{4}\boldsymbol{{n}}} }=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{2}{n}!}{{n}!.{n}!}\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\mathrm{2}^{\mathrm{4}{n}} }=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\frac{\mathrm{2}^{\mathrm{4}{n}} {n}!^{\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\left({n}!\right)^{\mathrm{4}} }}{\left(\mathrm{2}{n}−\mathrm{1}\right)\mathrm{2}^{\mathrm{4}{n}} } \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left({n}!\right)^{\mathrm{2}} }.\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}}+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right){n}!^{\mathrm{2}} }.\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right){n}!}.\frac{\left(\mathrm{1}\right)^{{n}} }{{n}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right)\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right){n}!}.\frac{\mathrm{1}^{{n}} }{{n}!} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}−\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \\ $$$${n}!=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+{k}\right)=\left(\mathrm{1}\right)_{{n}} \\ $$$${so}\:{We}\:{get}\:\:{S}=\frac{\mathrm{1}}{\mathrm{4}}+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} .−\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{\mathrm{2}\left(\mathrm{1}\right)_{{n}} }.\frac{\mathrm{1}^{{n}} }{{n}!} \\ $$$${recall}\:{that}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b};{c};{x}\right)=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}_{{n}} {b}_{{n}} }{{c}_{{n}} }\:\frac{{x}^{{n}} }{{n}!} \\ $$$$\Rightarrow{S}=\frac{\mathrm{1}}{\mathrm{4}}−\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{\left(\mathrm{1}\right)_{{n}} }=\frac{\mathrm{1}}{\mathrm{4}}−\left(\:\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};\mathrm{1}\right)−\mathrm{1}−\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{1}} }{\left(\mathrm{1}\right)_{\mathrm{1}} }.\frac{\mathrm{1}}{\mathrm{1}!}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}−\left(_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};\mathrm{1}\right)−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{1}−_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}};−\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};\mathrm{1}\right) \\ $$$${we}\:{use}\:{gausse}\:{Hypergeometrique}\:{theorem} \\ $$$$\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b};{c},\mathrm{1}\right)=\frac{\Gamma\left({c}\right)\Gamma\left({c}−{b}−{a}\right)}{\Gamma\left({c}−{a}\right)\Gamma\left({c}−{b}\right)}\:{this}\:{is}\:{just}\:{result} \\ $$$${of}\:\beta\left({b},{c}−{b}\right)_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b};{c};{z}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{b}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{c}−{b}+\mathrm{1}} \left(\mathrm{1}−{zx}\right)^{−{a}} {dx} \\ $$$${so}\:{we}\:{get} \\ $$$$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};\mathrm{1}\right)=\frac{\Gamma\left(\mathrm{1}\right)\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\Gamma\left(\mathrm{1}\right)\Gamma\left(\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}}{\pi} \\ $$$${S}=\mathrm{1}−_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};\mathrm{1}\right)=\mathrm{1}−\frac{\mathrm{2}}{\pi} \\ $$
Commented by mnjuly1970 last updated on 21/Sep/20
very nice thank you sir..
$${very}\:{nice}\:{thank}\:{you}\:{sir}.. \\ $$
Commented by maths mind last updated on 21/Sep/20
withe pleasur sir
$${withe}\:{pleasur}\:{sir} \\ $$

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