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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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    You're sooo close! Take a look at the limit definition of e (say at wikipedia) and just re-arrange the fractions inside the bracket.2011-11-09
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    No need to squeeze, you have got it. The radius of convergence is usually denoted by $r$, or $\rho$. You may need to use a source other than Wolfram Alpha for the limit assertion.2011-11-09
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    Thanks. Im really stuck, how would I explain/show how I found that lim $\frac{n^n}{(n+1)^n}$ = e?2011-11-09
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    It's $1/e$, not $e$. Flip the thing over. You get $\frac{(n+1)^n}{n^n}=\left(1+\frac{1}{n}\right)^n$. Does this look familiar? Some use the limit of this as the *definition* of $e$. It is a limit that I think can be found in most (all?) first-year calculus books.2011-11-10
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    Thanks! and yeap, sorry a typo!2011-11-10
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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 $\mathrm e$.