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Question Number 139276 by mnjuly1970 last updated on 25/Apr/21

Commented by mnjuly1970 last updated on 28/Apr/21

$$correctanswer::e^2.e^{e^2}$$

Answered by Dwaipayan Shikari last updated on 25/Apr/21

$$\varphi \left(m\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2^n}{\xi (m,n)}=m\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(2m\right)}^n}{n!}$$$$\xi (m,n)={\int }_0^{\mathrm{\infty }}{x}^n{e}^{-mx}dx=\frac{\mathrm{\Gamma }(n+1)}{m^{n+1}}$$$$\varphi \left(m\right)={me}^{2m}$$$$\mathrm{\Omega }=\underset{m=0}{\sum ^{\mathrm{\infty }}}\frac{(m+1)e^{(2m+2)}}{\mathrm{\Gamma }(m+2)}=e^2\underset{m=0}{\sum ^{\mathrm{\infty }}}\frac{e^{2m}}{m!}=e^2{e}^{e^2}$$

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