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Does the series $\displaystyle {\sum_{n=1}^{\infty}}\frac{n!}{n^n\,} \cdot e^n $ converge?

I tried the Ratio test. But because of the limit is $1$, this test does not give me any information about whether it converges or not.

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  • $\begingroup$ The limit of the ratio may be $1$, but since the ratio is always greater than $1$ this implies divergence $\endgroup$
    – Henry
    Nov 4, 2021 at 9:05

2 Answers 2

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From $e^n=\sum_{k=0}^\infty \frac{n^k}{k!}$ it follows that $e^n>\frac {n^n}{n!}$ (which is one of the positive summands). Hence in your series, each summand is $>1$.

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You could use the Stirling approximation: the term is asymptotic to $\sqrt{2\pi n}$, so a comparison tells us it diverges.

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