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Question Number 104471 by Dwaipayan Shikari last updated on 21/Jul/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{e^{n}n!}{n^n}$$

Answered by MAB last updated on 21/Jul/20

$$let{U}_n=\frac{n!}{n^n}{e}^n$$$$bystirlingformula:n!\backsim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$$$$wededuce:U_n\backsim \sqrt{2\pi n}$$$$hence\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{e^{n}n!}{n^n}diverges$$$$$$

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