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This question already has an answer here:

I want to determine the convergence of the following series: $$\Sigma_{n=1}^{\infty}\frac{n!}{n^n}i^n$$ I have tried applying the criteria of the n+1th over nth term: $$lim_{n->\infty}\left|\frac{(n+1)!}{(n+1)^{n+1}}i^{n+1}\frac{n^n}{n!i^n}\right|=lim_{n->\infty}\left|\frac{(n+1)n^n}{(n+1)^{n+1}}i\right|$$ $$=lim_{n->\infty}\left|\frac{n^n}{(n+1)^{n-1}}\right|$$ But how can I evaluate this limit?

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marked as duplicate by Martin R, Leucippus, Joshua Mundinger, José Carlos Santos complex-analysis Sep 3 at 6:04

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  • $\begingroup$ Try with its rediprocal... Besides you should start with $(n+1)^{n+1}$. $\endgroup$ – amsmath Sep 2 at 15:59
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Check your exponent on the denominator of $a_{n+1}.$ You should have

$\left| \dfrac{a_{n+1}}{a_n} \right|=\left|\dfrac{(n+1)!}{(n+1)^{n+1}}i^{n+1}\left(\dfrac{n^n}{n!i^n}\right)\right|$.

Then you are left with

$\lim_{n\to\infty}\left|\dfrac{(n+1)n^n}{(n+1)^{n+1}}\right|=\lim_{n\to\infty}\left(\dfrac{n}{n+1}\right)^n=\lim_{n\to\infty}\left(1-\dfrac{1}{n}\right)^n.$

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    $\begingroup$ The arrow can be realized by \to $\endgroup$ – amsmath Sep 2 at 16:02
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That is a nice series. By the Lagrange inversion theorem we have $$ W(x) = \sum_{n\geq 1}\frac{(-1)^{n+1}n^{n-1}}{n!}x^n $$ where $W$ is the Lambert function, i.e. the inverse of $xe^x$ in a neighbourhood of the origin. Since the only stationary point of $xe^x$ occurs at $x=-1$, the radius of convergence of the above series is $\frac{1}{e}$. This radius of convergence is unaffected by the substitution $x\mapsto -x$ or by differentiation. This leads to the fact that

$$ \sum_{n\geq 1}\frac{n^n}{n!}x^n = \frac{-W(-x)}{1+W(-x)} $$ holds for any $|x|<\frac{1}{e}$, and the LHS is the inverse function of $\frac{x}{x+1}e^{-\frac{x}{x+1}}$ in a neighbourhood of the origin.

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