# Finding convergence or divergence of a complex series [duplicate]

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?

## marked as duplicate by Martin R, Leucippus, Joshua Mundinger, José Carlos Santos complex-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 3 at 6:04

• Try with its rediprocal... Besides you should start with $(n+1)^{n+1}$. – amsmath Sep 2 at 15:59

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

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

• The arrow can be realized by \to – amsmath Sep 2 at 16:02

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.