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Im trying to find the radius of convergence for $\sum_{n=0}^\infty \frac{n!}{n^n}z^n.$

Applying the ratio test $\frac{C_{n+1}}{C_n}$, I simplified $\frac{n!}{n^n}$ to $\frac{n^n}{(n+1)^n}$.

On Wolfram, it says the limit of $\frac{n^n}{(n+1)^n}$ as $n$ tends to infinity is $\frac{1}{e}$.

Im trying to find $ \rho = \frac{1}{\lim \limits_{n \to \infty} \frac{C_{n+1}}{C_n}}, $

so $\rho = e$?

If not, how would I go about finding the radius of convergence? Do I need to use the squeeze theorem maybe? Thanks.

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    This is a standard Calculus question meant to discover whether the student recognizes or recollects the definition of $e$.2011-11-25

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The question is resolved in the comments. To summarize the argument, we have $ \rho = \left( \lim \limits_{n \to \infty} \frac{n^n}{(n+1)^n} \right)^{-1} = \lim \limits_{n \to \infty} \frac{(n+1)^n}{n^n} = \lim \limits_{n \to \infty} \Big(1+\frac{1}{n} \Big)^n = \mathrm e, $ just from the definition of $\mathrm e$.