observation from a question on m.se

Set g(z)=\sum_{n=1}^{\infty} \left(\frac{z}{n}\right)^{n}, then the sum:

\sum_{m=1}^{\infty} \frac{m!}{m^{m}} can be interpreted as the value of the Laplace

Transform of g(z) at s=1:

\sum_{m=1}^{\infty} \frac{m!}{m^{m}} =  \sum_{m=1}^{\infty} \frac{1}{m^{m}} \int_{0}^{\infty} t^{m}e^{-t} dt = \int_{0}^{\infty} e^{-t} \sum_{m=1}^{\infty} \frac{t^{m}}{m^{m}} dt

 

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