$\sum_{n=1}^\infty \frac{n^n}{e^nn!}$发散

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相邻比值趋于1，比值审敛法无法判断
SumConvergence[n^n/E^n/n!, n]

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类似题
根据Stirling存在常数$c,C$
\[\frac{c}{\sqrt{n}}\le \frac{(\frac{n}{e})^n}{n!}\le \frac{C}{\sqrt{n}}.\]
因为$\sum_n \frac{1}{\sqrt{n}}$发散，所以原级数发散。

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huggingface.co/spaces/Qwen/Qwen2.5-Math-Demo
To determine the divergence of the series \(\sum_{n=1}^{\infty} \frac{e^{-n} n^n}{n!}\), we can use Stirling's approximation for factorials. Stirling's approximation states that for large \(n\),
\[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n. \]
Using this approximation, we can rewrite the general term of the series as follows:
\[ \frac{e^{-n} n^n}{n!} \approx \frac{e^{-n} n^n}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n} = \frac{e^{-n} n^n}{\sqrt{2\pi n} \cdot \frac{n^n}{e^n}} = \frac{e^{-n} n^n \cdot e^n}{\sqrt{2\pi n} \cdot n^n} = \frac{1}{\sqrt{2\pi n}}. \]
Now, we need to consider the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{2\pi n}}\). This series is a constant multiple of the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\), which is a \(p\)-series with \(p = \frac{1}{2}\). A \(p\)-series \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) diverges if \(p \leq 1\). Since \(p = \frac{1}{2} \leq 1\), the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) diverges, and therefore the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{2\pi n}}\) also diverges.

Thus, the original series \(\sum_{n=1}^{\infty} \frac{e^{-n} n^n}{n!}\) diverges.