# Euler's divergent series sum n!*(-1)^n: what is known about the resulting constant?

Much of the theory of continued fractions has been developped by Euler in the 18th century. The little survey "Euler: continued fractions and divergent series (and Nicholas Bernoulli)", mentions towards the end the continued fraction $$f(x)=\dfrac{1}{1+\dfrac{x}{1+\dfrac{x}{1+\dfrac{2x}{1+\dfrac{2x}{1+\dfrac{3x}{1+\dots}}}}}}$$ which Euler has "derived" from the divergent power series $1-x+2x^2-6x^3+24x^4-+...$ .

For $x=1$, the continued fraction converges to a limit $f(1)\approx 0.5963475922$. I was wondering: what is known about the values $f(x)$, in particular $f(1)$? Are they known to be transcendental for $x\in\mathbb N$? Can they be expressed by other known constants?

• The standard formal manipulation $$\sum_{n=0}^\infty (-1)^n n! = \sum_{n=0}^\infty (-1)^n \int_0^\infty t^n e^{-t} dt = \int_0^\infty \left(\sum_{n=0}^\infty (-t)^n\right) e^{-t} dt = \int_0^\infty \frac{e^{-t} dt}{1+t}$$ suggests that we're dealing with the "Gompertz constant" $-e \phantom. {\rm Ei}(-1) = .596347362\ldots$ (see the Mathworld entry and its link to OEIS A073003). This is not quite the decimal expansion you give. – Noam D. Elkies Jan 14 '12 at 18:32
• Thank you. Long live the "formal manipulations"! Now: Is there a way to see that the above continued fraction is equal to the integral $\int_0^\infty \frac{e^{-t} dt}{1+tx}$ without using a divergent sum? – Wolfgang Jan 14 '12 at 19:05
• @Spanferkel: You're welcome, and good question. I don't have a proof, but you might be able to construct one by extending $1 = \int_0^\infty e^{-t} dt$ and $\int_0^\infty e^{-t} dt/(1+xt)$ to an infinite sequence of definite integrals and establishing a linear recurrence. cf. Henry Cohn's "A Short Proof of the Simple Continued Fraction Expansion of $e$" in The American Mathematical Monthly, Vol. 113 #1 (Jan.2006), 57–62. – Noam D. Elkies Jan 14 '12 at 19:32

The sequence $a_n=(-1)^n n!$ satisfies $a_{n+1}+(n+1)a_n = 0$ (with $a(0)=1$). Thus the generating function satisfies the differential equation $x^2y'+(x+1)y=1$ (where $y(0)=1$). The unique solution is $$\frac{e^{\frac{1}{x}}Ei\left(1,\frac{1}{x}\right)}{x}$$ For $x=1$, the constant is $e Ei(1,1) \approx .5963473623231940743410785$.

The reason this is justified is because the sequence is Gevrey, i.e. it does not diverge 'too fast', so associated to it is a unique (generating) function which has that sequence as coefficients (asymptotically) at 0, when approached along the real line. The modern theory that 'this all works' is essentially due to Écalle, although Lindelof had worked out quite a bit already $100$ years before.

• Edited only to fix a typo ("$n+1$" was "$n+!$"). I don't know this theory; does it also identify the solution $y(x) = x^{-1} e^{1/x} {\rm Ei}(1,x^{-1})$ of that differential equation with Euler's continued fraction? – Noam D. Elkies Jan 14 '12 at 21:01
• My knowledge of the theory only extends to sequences and asymptotics of functions. I would say that it really depends on what kind of equation one can directly derive from the continued fraction, but that is really just speculation. – Jacques Carette Jan 15 '12 at 0:42
• It's a P-finite sequence. Such DEs have holonomic functions as solutions. Read everything from Zeilberger. – rwst Feb 15 '14 at 14:53

Not an answer per se, but check out this nice article. (How Euler did it, by Ed Sandifer)

A small - surely not authoritative- bit of information: http://www.groupsrv.com/science/about477640.html

More facts and references at https://oeis.org/A073003, not the least of which is that it appears in Ramanujan's notebooks.

A nice sequence of rational numbers approximating the constant $f(1)$ is $$s(n) = \sum_{k=0}^{n-1} {(n+1-k)^k \over (n+2-k)^{1+k} }$$ and $$A = f(1) = \lim_{n \to \infty} s(n)$$ The $s(n)$ occur here as partial sums of a series which was derived by the analysis of the triangle of "Eulerian numbers"
The first five approximants are

  [1/2, 7/12, 43/72, 323/540, 77411/129600,...]


and in steps of $10$ the partial sums are

  s( 0):     0.500000000000
s(10):     0.596329104980
s(20):     0.596347376719
s(30):     0.596347363391
s(40):     0.596347362340
s(50):     0.596347362311
s(60):     0.596347362325
s(70):     0.596347362323
s(80):     0.596347362323


(See a discussion at this question in MSE )

Regarding the expression of $$f(1)$$ in terms of other known constants: $$f(1) = -e\left(\gamma + \sum_{n\geq 1} (-1)^n \frac1{n\cdot n!} \right) \qquad\quad \tag{*} \label{id}$$ $$\qquad\qquad\qquad= -e \left(\gamma - 1 + \frac1{4} - \frac1{18} + \frac1{96} - \frac1{600} + \cdots \right)$$ where $$e = \sum_{n\geq0} \frac1{n!} \approx 2.718$$ and $$\gamma \approx 0.577$$ is the Euler-Mascheroni constant.

This infinite sum expression follows from the identity $$f(1) = -e {\rm Ei}(-1)$$ and is on the Wikipedia page for the Gompertz constant. I recommend also taking a look at the survey Euler's constant: Euler's work and modern developments by Jeffrey Lagarias, where it appears as (2.5.11).

Since $$f(x) =" \sum_{n\geq 0} (-1)^n n! x^n$$ and $$e^{-1} = \sum_{n\geq 0} (-1)^n \frac{1}{n!}$$, the identity ($$*$$) can be expressed more symmetrically as $$- \left(\sum_{n\geq 0} (-1)^n n! \right) \left(\sum_{m\geq 0} (-1)^m \frac{1}{m!} \right) \quad =" \quad \gamma + \sum_{n\geq 1} (-1)^n \frac1{n\cdot n!}.$$

This formula suggests an argument for evaluating $$f(1)$$ which avoids differential equations, if one is willing to make some convenient cancellations of divergent series after rearranging the two-index summation on the left-hand side.

For arbitrary nonzero $$x$$, the relation ($$*$$) generalizes to $$f(x) = - \frac1{x} e^{1/x} {\rm Ei}(-\frac1{x})$$ where $${\rm Ei}(x)$$ denotes the exponential integral $${\rm Ei}(x) = \int_{-\infty}^x \frac{e^t}{t} dt.$$ Using the Taylor expansion of $${\rm Ei}(x)$$, $$f(x) = - \frac1{x} e^{1/x} \left(\gamma + \ln x + \sum_{n\geq 1} \frac{(-1)^n}{n \cdot n!} \frac1{ x^n} \right).$$

• Thank you. I like the $="$ notation. :) – Wolfgang Sep 23 at 6:49