# 1 − 1 + 2 − 6 + 24 − 120 + ...

In mathematics, the divergent series

${\displaystyle \sum _{k=0}^{\infty }(-1)^{k}k!}$

was first considered by Euler, who applied summability methods to assign a finite value to the series.[1] The series is a sum of factorials that alternatingly are added or subtracted. A way to assign a value to the divergent series is by using Borel summation, where one formally writes

${\displaystyle \sum _{k=0}^{\infty }(-1)^{k}k!=\sum _{k=0}^{\infty }(-1)^{k}\int _{0}^{\infty }x^{k}e^{-x}\,dx}$

If summation and integration are interchanged (ignoring that neither side converges), one obtains:

${\displaystyle \sum _{k=0}^{\infty }(-1)^{k}k!=\int _{0}^{\infty }\left[\sum _{k=0}^{\infty }(-x)^{k}\right]e^{-x}\,dx}$

The summation in the square brackets converges and equals 1/1 + x if |x| < 1. If we analytically continue this 1/1 + x for all real x, one obtains a convergent integral for the summation:

{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }(-1)^{k}k!&=\int _{0}^{\infty }{\frac {e^{-x}}{1+x}}\,dx\\[4pt]&=eE_{1}(1)\approx 0.596\,347\,362\,323\,194\,074\,341\,078\,499\,369\ldots \end{aligned}}}

where E1(z) is the exponential integral. This is by definition the Borel sum of the series.

## Derivation

Consider the coupled system of differential equations

${\displaystyle {\dot {x}}(t)=x(t)-y(t),\qquad {\dot {y}}(t)=-y(t)^{2}}$

where dots denote derivatives with respect to t.

The solution with stable equilibrium at (x,y) = (0,0) as t → ∞ has y(t) = 1/t, and substituting it into the first equation gives a formal series solution

${\displaystyle x(t)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {(n-1)!}{t^{n}}}}$

Observe x(1) is precisely Euler's series.

On the other hand, the system of differential equations has a solution

${\displaystyle x(t)=e^{t}\int _{t}^{\infty }{\frac {e^{-u}}{u}}\,du.}$

By successively integrating by parts, the formal power series is recovered as an asymptotic approximation to this expression for x(t). Euler argues (more or less) that setting equals to equals gives

${\displaystyle \sum _{n=1}^{\infty }(-1)^{n+1}(n-1)!=e\int _{1}^{\infty }{\frac {e^{-u}}{u}}\,du.}$