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 $\mathrm{e}^{-1} (\mathrm{Ei}(1) + i\pi)$.

  • 2
    $\begingroup$ If you use the method of Borel summation you will come to the same result, you have only to pay attention to the fact that the exponential integral is a multivaluated function so you have to be careful. $\endgroup$ – Renato Faraone Apr 19 '16 at 16:09
  • $\begingroup$ @RenatoFaraone I see, thanks! I assume the correct answer is $0.697\ldots$, then? $\endgroup$ – user76284 Apr 19 '16 at 16:36
  • 1
    $\begingroup$ Yes, if we want to assign a value to a divergent series through a function we'll probably prefer to choose it's principal branch value. $\endgroup$ – Renato Faraone Apr 19 '16 at 16:57
  • $\begingroup$ Is it sure, that Borel-summation can be applied here? I was convinced, that to be Borel-summable a series must have alternating signs (if I understood Konrad Knopp's book correctly) $\endgroup$ – Gottfried Helms Apr 25 '16 at 18:03

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