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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?

flag	asked Jan 14 2012 at 18:13 spanferkel 883●2●9

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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 2012 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? – spanferkel Jan 14 2012 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 2012 at 19:32

Noam D. Elkies · Accepted Answer · 2012-01-14 20:58:45Z UTC

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.

Igor Rivin · Answer 2 · 2012-01-14 18:29:55Z UTC

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

answered Jan 14 2012 at 18:29

Igor Rivin
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Gottfried Helms · Answer 3 · 2012-01-14 19:18:03Z UTC

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

answered Jan 14 2012 at 19:18

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

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