What does wolfram alpha means by this $$\lim_{n \to \infty} \frac{n!(-e)^n}{n^n} = \exp (i~0~to~2\pi)\infty$$
Does it have a specific meaningful answer or is it just infinity? is it related to complex series? where can I read more about it?
 A: The reason for this starts with the two limits
$$ \lim_{\substack{n \to \infty \\ n \text{ even}}} \frac{n!(-e)^n}{n^n} = +\infty \text{ and } \lim_{\substack{n \to \infty \\ n \text{ odd}}} \frac{n!(-e)^n}{n^n} = -\infty. $$
Notice that the $(-1)^n$ in the numerator causes it to flip between positive and negative.
Now what I believe WolframAlpha is doing is treating $n$ as a real number. If this happens then
$$ (-1)^n = (e^{\pi i})^n = e^{n \pi i}. $$
Now, if you look at the sequence $n = a, a + 2\pi, a + 4\pi, \dots, a+ 2k\pi,\dots$ this value is always equal to
$$ e^{a \pi i}. $$
The rest of the limit, $n!e^n/n^n$ goes to infinity, so what you are left with is
$$ e^{a \pi i} \infty $$
where $a$ can take on any real number. But because adding or subtracting $2\pi$ from $a$ doesn't change the value of the exponential, we can restrict our attention to $a$ in the interval $[0,2\pi)$. This gives the answer
$$ e^{[0,2\pi)i} \infty $$
which is supposed to represent a value of $\infty$ but at every possible angle in the complex plane.
