Consider the series
$$\sum_{n=0}^\infty n! = 0! + 1! + 2! + 3! + 4! + \ldots = 1 + 1 + 2 + 6 + 24 + \ldots$$
This series clearly diverges. Now, given that the Gamma function is defined by
$$n! = \Gamma(n+1) = \int_0^\infty t^n \mathrm{e}^{-t} \mathrm{d}t$$
we obtain
$$\sum_{n=0}^\infty n! = \sum_{n=0}^\infty \int_0^\infty t^n \mathrm{e}^{-t} \mathrm{d}t$$
Interchanging limits for the sake of regularization yields
\begin{align} \sum_{n=0}^\infty n! &= \int_0^\infty \sum_{n=0}^\infty t^n \mathrm{e}^{-t} \mathrm{d}t \\ &= \int_0^\infty \frac{\mathrm{e}^{-t}}{1 - t} \mathrm{d}t \\ &= \left[ -\mathrm{e}^{-1} \mathrm{Ei}(1-t) \right]_0^\infty \\ &= \mathrm{e}^{-1} \mathrm{Ei}(1) \\ &\approx 0.697175 \end{align}
where $\mathrm{Ei}$ is the exponential integral. Notice that we used the Cauchy principal value since the function has a pole at $t = 1$. Is this a valid regularization of the series?
Entering the command
N@Sum[n!, {n, 0, Infinity}, Regularization -> "Borel"]
into Mathematica yields
0.697175 + 1.15573 I
which is a complex value. Why is the answer different?
I know this is a divergent series, but I would like to find its regularized value.
EDIT: 0.697175 + 1.15573 I is just $\mathrm{e}^{-1} (\mathrm{Ei}(1) + i\pi)$.