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1 − 1 + 2 − 6 + 24 − 120 + ... : ウィキペディア英語版
1 − 1 + 2 − 6 + 24 − 120 + ...
In mathematics, the divergent series
:\sum_^\infty (-1)^k k!
was first considered by Euler, who applied summability methods to assign a finite value to the series.〔〕 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 we formally write
:\sum_^\infty (-1)^k k! = \sum_^\infty (-1)^k \int_0^\infty x^k \exp(-x) \, dx
If we interchange summation and integration (ignoring the fact that neither side converges), we obtain:
:\sum_^\infty (-1)^k k! = \int_0^\infty \left((-x)^k \right )\exp(-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'', we obtain a convergent integral for the summation:
:\sum_^\infty (-1)^ k! = \int_0^\infty \frac \, dx = e E_1 (1) \approx 0.596347362323194074341078499369\ldots
where E_1 (z) is the exponential integral. This is by definition the Borel sum of the series.
==Derivation==

Consider the coupled system of differential equations
:\dot(t) = x(t) - y(t),\qquad \dot(t)=-y(t)^
where dots denote time derivatives.
The solution with stable equilibrium at (x,y)=(0,0) as t\to\infty has y(t)=1/t. And substituting it into the first equation gives us a formal series solution
:x(t) = \sum^_(-1)^\frac\int^_\frac\mathrmu.
By successively integrating by parts, we recover the formal power series as an asymptotic approximation to this expression for x(t). Euler argues (more or less) that setting equals to equals gives us
: \sum^_(-1)^(n-1)! = e\int^_\frac\mathrmu.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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