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Question Number 98434 by  M±th+et+s last updated on 14/Jun/20

$$provethat$$$$\mathrm{\Omega }=\underset{n=0}{\sum ^{+\mathrm{\infty }}}\underset{m=0}{\sum ^{+\mathrm{\infty }}}\frac{\mathrm{\Gamma }(n+\frac{1}{2}).\mathrm{\Gamma }(m+\frac{1}{2})}{\mathrm{\Gamma }(n+1).\mathrm{\Gamma }(m+1)}.\frac{{\left(\frac{2}{3}\right)}^n.{\left(\frac{1}{2}\right)}^m}{(n+m+\frac{1}{2})}$$$$=\frac{\sqrt{3}}{2\sqrt{2}}.G_{2,2}^{2,2}\left(\frac{1}{4}{\mid }_{0,0}^{\frac{1}{2},\frac{1}{2}}\right)$$$$=\frac{\sqrt{3}\pi }{\sqrt{2}}K\left(\frac{3}{4}\right)=\frac{\sqrt{3}{\pi }^2}{2\sqrt{2}AGM(1,\frac{1}{2})}$$

Commented by  M±th+et+s last updated on 13/Jun/20

$$G_{2,2}^{2,2}(Z\mid \bullet )\equiv meijerG-function\equiv \left(specialhighfunction\right)$$$$K\left(m\right)isthecompleteellipticintegraloffirstkind$$$$withparameterm=k^2$$$$\mathrm{\Gamma }\left(s\right)isgammafunction$$$$AGM(x,y)istheArithmetic-geometricmeanofxandy.$$$$$$

Commented by maths mind last updated on 14/Jun/20

$$\mathrm{\Omega }={\left(\sum _{n⩾0}\frac{\mathrm{\Gamma }(n+\frac{1}{2})}{\mathrm{\Gamma }(n+1)}\right)}^2$$$$\frac{\mathrm{\Gamma }(n+\frac{1}{2})}{\mathrm{\Gamma }(n+1)}⩾\frac{\mathrm{\Gamma }\left(n\right)}{\mathrm{\Gamma }(n+1)}=\frac{1}{n}$$$$\Rightarrow \sum _{n⩾0}\frac{\mathrm{\Gamma }(n+\frac{1}{2})}{\mathrm{\Gamma }(n+1)}⩾\sum _{n⩾1}\frac{1}{n}...Sum/Diverge$$$$chekitsir$$

Commented by  M±th+et+s last updated on 14/Jun/20

$$thankyousir.$$$$youarerighsiricheckedittherewere$$$$sometypos$$$$\ast \ast iamsorry\ast \ast$$

Commented by maths mind last updated on 14/Jun/20

$$thankyouforthisniceQuation$$

Commented by  M±th+et+s last updated on 14/Jun/20

$$iamgladyoulikedit$$

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