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Question Number 128122 by Dwaipayan Shikari last updated on 04/Jan/21

1+(1/(16))+(5^2 /(16^2 .2!))+((5^2 .9^2 )/(16^3 .3!))+((5^2 .9^2 .13^2 )/(16^4 .4!))+...=((√π)/(Γ^2 ((3/4))))=F_1 ((1/4),(1/4),1;1) Prove The above relation Where F_1 (Φ,ϕ,γ;μ)=Σ_(n≥0) ^∞ (((Φ)_n (ϕ)_n )/(n!(γ)_n ))μ^n (ζ)_n =ζ(ζ+1)(ζ+2)...(ζ+n−1)

$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{2}} .\mathrm{2}!}+\frac{\mathrm{5}^{\mathrm{2}} .\mathrm{9}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{3}} .\mathrm{3}!}+\frac{\mathrm{5}^{\mathrm{2}} .\mathrm{9}^{\mathrm{2}} .\mathrm{13}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{4}} .\mathrm{4}!}+...=\frac{\sqrt{\pi}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)}={F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}},\mathrm{1};\mathrm{1}\right) \\ $$$${Prove}\:{The}\:{above}\:{relation} \\ $$$${Where} \\ $$$${F}_{\mathrm{1}} \left(\Phi,\varphi,\gamma;\mu\right)=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\Phi\right)_{{n}} \left(\varphi\right)_{{n}} }{{n}!\left(\gamma\right)_{{n}} }\mu^{{n}} \\ $$$$\left(\zeta\right)_{{n}} =\zeta\left(\zeta+\mathrm{1}\right)\left(\zeta+\mathrm{2}\right)...\left(\zeta+{n}−\mathrm{1}\right) \\ $$

Answered by mindispower last updated on 05/Jan/21

=1+Σ_(n≥1) ((Π_(k=0) ^(n−1) (4k+1)^2 )/(16^n (n!)^2 )) =1+Σ_(n≥1) ((4^(2n) .Π_(k=0) ^(n−1) (k+(1/4)).Π(k+(1/4)))/(.16^n .n!)).(1/(n!)) =1+Σ_(n≥1) ((((1/4))_n .((1/4))_n )/((1)_n )).(1^n /(n!))=_2 F_1 ((1/4),(1/4);1;[1]) than use relation _2 F_1 (a,b,c;1) withe β, Γ

$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{4}{k}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{16}^{{n}} \left({n}!\right)^{\mathrm{2}} }\:\: \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{4}^{\mathrm{2}{n}} .\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{4}}\right).\Pi\left({k}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{.\mathrm{16}^{{n}} .{n}!}.\frac{\mathrm{1}}{{n}!} \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)_{{n}} .\left(\frac{\mathrm{1}}{\mathrm{4}}\right)_{{n}} }{\left(\mathrm{1}\right)_{{n}} }.\frac{\mathrm{1}^{{n}} }{{n}!}=_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}};\mathrm{1};\left[\mathrm{1}\right]\right) \\ $$$${than}\:{use}\:{relation}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b},{c};\mathrm{1}\right)\:{withe}\:\beta, \\ $$$$\Gamma\: \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 05/Jan/21

Great sir ! then _2 F_1 (a,b,c;1)=((Γ(c)Γ(c−a−b))/(Γ(c−a)Γ(c−b))) _2 F_1 ((1/4),(1/4);1;1)=((Γ((1/2)))/(Γ^2 ((3/4))))=((√π)/(Γ^2 ((3/4)))) Is it sir??

$${Great}\:{sir}\:! \\ $$$${then}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b},{c};\mathrm{1}\right)=\frac{\Gamma\left({c}\right)\Gamma\left({c}−{a}−{b}\right)}{\Gamma\left({c}−{a}\right)\Gamma\left({c}−{b}\right)}\:\: \\ $$$$\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}};\mathrm{1};\mathrm{1}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)}=\frac{\sqrt{\pi}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$${Is}\:{it}\:\:{sir}?? \\ $$

Commented by mindispower last updated on 06/Jan/21

yes sir great

$${yes}\:{sir}\:{great} \\ $$

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