Discussion
The radius of convergence of the power series \(\displaystyle\sum_{n=1}^{\infty} \frac{n! z^{n}}{n^{n}}\) is :
Answer: B
The radius of convergence is \(e\).
With \(a_{n} = n!/n^{n}\), the ratio test gives
\[ \begin{aligned} \frac{a_{n}}{a_{n+1}} &= \frac{n!}{n^{n}} \cdot \frac{(n+1)^{n+1}}{(n+1)!} \\ &= \frac{(n+1)^{n}}{n^{n}} = \left(1 + \frac{1}{n}\right)^{n} \to e \end{aligned} \]
so \(R = \lim |a_{n}/a_{n+1}| = e\). Stirling's formula \(n! \approx \sqrt{2\pi n}\,(n/e)^{n}\) confirms it: \(|a_{n}|^{1/n} \to 1/e\).
The option \(1/e\) is the limit of \(|a_{n}|^{1/n}\), not its reciprocal. The radius is not 1, because the factorial and the power do not cancel, and not \(\infty\), because \(n!/n^{n}\) decays only like \(e^{-n}\) (times \(\sqrt{2\pi n}\)), not faster than every geometric rate.
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