Question Number 98434 by M±th+et+s last updated on 14/Jun/20
$${prove}\:{that} \\ $$$$\Omega=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\underset{{m}=\mathrm{0}\:} {\overset{+\infty} {\sum}}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left({m}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right).\Gamma\left({m}+\mathrm{1}\right)}.\frac{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} .\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{m}} }{\left({n}+{m}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}}.{G}_{\mathrm{2},\mathrm{2}} ^{\mathrm{2},\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\mid_{\mathrm{0}\:\:,\:\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$$=\frac{\sqrt{\mathrm{3}}\pi}{\:\sqrt{\mathrm{2}}}{K}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{3}}\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}{AGM}\left(\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$
Commented by M±th+et+s last updated on 13/Jun/20
$${G}_{\mathrm{2},\mathrm{2}} ^{\mathrm{2},\mathrm{2}} \left({Z}\mid\bullet\right)\equiv{meijer}\:{G}−{function}\equiv\left({special}\:{high}\:{function}\right) \\ $$$${K}\left({m}\right)\:{is}\:{the}\:{complete}\:{elliptic}\:{integral}\:{of}\:{first}\:{kind} \\ $$$${with}\:{parameter}\:{m}={k}^{\mathrm{2}} \\ $$$$\Gamma\left({s}\right)\:{is}\:{gamma}\:{function} \\ $$$${AGM}\left({x},{y}\right)\:{is}\:{the}\:{Arithmetic}−{geometric}\:{mean}\:{of}\:{x}\:{and}\:{y}. \\ $$$$ \\ $$
Commented by maths mind last updated on 14/Jun/20
$$\Omega=\left(\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}\right)^{\mathrm{2}} \\ $$$$\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}\geqslant\frac{\Gamma\left({n}\right)}{\Gamma\left({n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{n}} \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}\geqslant\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}…{Sum}/{Diverge} \\ $$$${chek}\:{it}\:{sir}\: \\ $$
Commented by M±th+et+s last updated on 14/Jun/20
$${thank}\:{you}\:{sir}. \\ $$$${you}\:{are}\:{righ}\:{sir}\:{i}\:{checked}\:{it}\:{there}\:{were} \\ $$$${some}\:{typos}\: \\ $$$$\ast\ast{iam}\:{sorry}\ast\ast \\ $$
Commented by maths mind last updated on 14/Jun/20
$${thank}\:{you}\:{for}\:{this}\:{nice}\:{Quation} \\ $$
Commented by M±th+et+s last updated on 14/Jun/20
$$\:{i}\:{am}\:{glad}\:{you}\:{liked}\:{it} \\ $$