Can the transcendence of e be shown using its continued fraction expansion e = [2;1,2,1,1,4,1,1,6,...]?

5$\begingroup$ The following paper by Henry Cohn at least demonstrates some relation between the transcendence of $e$ and its continued fraction expansion: arxiv.org/PS_cache/math/pdf/0601/0601660v3.pdf $\endgroup$ – Pete L. Clark May 17 '10 at 1:38

13$\begingroup$ "Interesting read." Well, yes  I said it was by Henry Cohn, so my statement implies yours. :) $\endgroup$ – Pete L. Clark May 17 '10 at 3:54

$\begingroup$ BTW, can anyone show a reference where I can find the proof that e=[2;1,2,11,4,11,6,1,1,...]? $\endgroup$ – TCL Mar 7 '11 at 17:35

$\begingroup$ @TCL: I hadn't seen this thread before, otherwise I would have answered. I discuss te continued fraction of $e$ at length in a blog post here: topologicalmusings.wordpress.com/2008/08/04/… Lots of references are given, and there is some useful discussion by Henry Cohn in the comments. $\endgroup$ – Todd Trimble♦ Oct 19 '11 at 10:33

$\begingroup$ @TCL Wolfram Alpha shows e=[2;1,2,1,1,4,1,1,6,1,1,...], but see here for more representations. $\endgroup$ – Mr Pie Jun 28 at 5:04
I have to join Gerry in his claim that $e$ is uniquely determined by its continued fraction, but transcendence proofs for the latter use the number $e$ instead. Without pretending to prove the transcendence of $e$ (proofs can be found in so many books and articles), I would like to mention that more general "pseudoperiodic" continued fractions of the form $\alpha=[b_0;b_1,\dots,b_s,(\overline{c_1+\lambda d_1,\dots,c_m+\lambda d_m})^\infty_{\lambda=0}]$ are known to be transcendental. This is because they can be expressed through the values of socalled Siegel's $E$functions at rational points, and the transcendence result for the latter goes back to Siegel's 1929 paper.
The result where the use of continued fraction for $e$ is crucial is the following [C.S. Davis, Bull. Austral. Math. Soc. 20 (1979) 407410]. For each $C<1/2$, the inequality $$\biggle\frac pq\biggr< C\frac{\log\log q}{q^2\log q}$$ has only finitely many solutions in rationals $p/q$, and at the same time there are infinitely many solutions of the inequality if one takes $C>1/2$. This means that the continued fraction gives us an efficient way to measure the "quality" of rational approximations to $e$, and this is exactly the thing continued fractions are designed for! This result was generalized to the general "pseudoperiodic" continued fractions in [B.G. Tasoev, Math. Notes 67 (2000) 786791].
I know only one family of examples where continued fractions are used directly to prove the transcendence of the numbers they represent. This is related to Mahler's method and the numbers are Liouvillian numbers, so that they are "too good" approximated by rationals.
In his article Transcendental continued fractions, Journal of Number Theory 13, November 1981, 456462, Gideon Nettler shows that two numbers given by continued fractions $A = [a_0,a_1,a_2,...]$ and $B = [b_0, b_1, b_2, ...]$ have the property that $A$, $B$, $A \pm B$, $A/B$ and $AB$ are irrational if $\frac12 a_n > b_n > a_{n1}^{5n}$ for sufficiently large $n$, and transcendental if $a_n > b_n > a_{n−1}^{(n−1)^2}$ for sufficiently large $n$. The growth of the $a_i$ in the continued fraction expansion of $e$ is so small that present methods seem useless for proving the transcendence of $e$ in this way.
Edit. Similarly, Alan Baker proved in Continued fractions of transcendental numbers (Mathematika 9 (1962), 18) that if $q_n$ denotes the denominator of the $n$th convergent of a continued fraction $A$, and if $$ \lim \sup \frac{(\log \log q_n)(\log n)^{1/2}}{n} = \infty, $$ then $A$ is transcendental.
Edit 2 I guess that the answer to your question should be a firm "yes" after all. In
 Über einige Anwendungen diophantischer Approximationen, Abh. Preu\ss. Akad. Wiss. 1929; Gesammelte Abhandlungen, vol I, p. 209241
Siegel proved that all continued fractions $$ \frac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \ldots}}} $$ in which the $a_i$ form a nonconstant arithmetic sequence are transcendental. Applying this to the continued fraction expansion of $\frac{e1}{e+1}$ gives the transcendence of $e$.
Siegel's proof uses, predictably, analytic machinery (solutions of Bessel and Riccati differential equations) going far beyond Liouville's theorem.

$\begingroup$ Might it not be better to say that it has been proved that various Bessel functions take on transcendental values at integer or rational arguments, and that among these transcendental Bessel values are some which have partial quotients in arithmetic progression? $\endgroup$ – Gerry Myerson Oct 17 '10 at 10:03

$\begingroup$ I don't think so. Euler computed the value of continued fractions in which the $a_i$ form an arithmetic sequence in terms of solutions of the Riccati equation (E71 on the Euler archive). This means there's a direct way from the continued fraction to the differential equation. $\endgroup$ – Franz Lemmermeyer Oct 18 '10 at 7:58

$\begingroup$ Franz, commenting on your Edit 2: Siegel indeed settled the grounds of a new method in transcendental number theory (generalising LindemannHermiteWeierstrass rather than Liouville). However, this method makes no use of continued fraction expansions. Quite opposite, the fact that "pseudoperiodic" CFs are transcendental followed from Siegel's general theorem on the transcendence of the values of Bessel functions at nonzero algebraic points. $\endgroup$ – Wadim Zudilin Dec 15 '10 at 11:08

$\begingroup$ @Wadim: yes. But you can rewrite Siegel's proof as follows: start with the periodic CFs, show that they satisfy the Bessel differential equation, and finally prove that the Bessel functions have transcendental values at nonzero algebraic points. $\endgroup$ – Franz Lemmermeyer Dec 15 '10 at 11:56
There is a notveryhelpful sense in which the answer is yes: the continued fraction expansion determines $e$, and therefore can be used to prove anything one can prove about $e$, including its transcendence.
We know so little about the continued fractions of real algebraic numbers of degree exceeding 2 that I would be surprised if there were any direct path from continued fraction to transcendence.

13$\begingroup$ @Gerry  Actually there has been quite a bit of work done on the transcendence of continued fractions. The first transcendence criteria for continued fractions were proved by E. Maillet, H. Davenport and K.F. Roth, A. Baker, and more recently improved by B. Adamczewski and Y. Bugeaud. For instance, see: math.univlyon1.fr/homeswww/adamczew/ABD.pdf and references therein. A particularly nice result is the following: if r is a real number with an infinite continued fraction expansion that begins with infinitely many palindromes, then r is either quadratic irrational or transcendental. $\endgroup$ – Amy Glen Jul 19 '10 at 20:21
To use the continued fraction for a number to prove it's transcendental, one usually shows that the rational approximations it affords are "too good" for an algebraic number. The traditional tool here is Liouville's theorem, but this has been improved to the more powerful Roth's theorem:
If $\alpha$ is algebraic, then for every $\varepsilon > 0$ there are only finitely many rational numbers $p/q$ satisfying $$ \left\alpha  \frac{p}{q}\right < \frac{1}{q^{2+\epsilon}}. $$
Unfortunately, $e$ also satisfies this. Indeed, rational approximations to $e$ are uncommonly bad. As Wadim mentions, there are only finitely many $p/q$ satisfying the following inequality. $$ \lefte  \frac{p}{q}\right < \frac{\log \log q}{3 q^2 \log q}. $$
But Khinchin proved that, for almost all $\alpha$, $$ \left\alpha  \frac{p}{q}\right < \frac{1}{q} \phi(q) $$ has infinitely many solutions if and only if $\sum_q \phi(q)$ diverges.
Additionally, Khinchin showed that the geometric means of the entries of the continued fraction expansion of a real number almost always converge to a universal constant. The geometric means of the entries of the continued fraction expansion of $e$ diverge.
If either of Khinchin's conditions hold for nonquadratic algebraic numbers, the transcendentality of $e$ would follow, but a proof is probably out of reach.
Finally, the continued fraction expansion of $e$ provides an immediate proof of its irrationality, because rational numbers have finite continued fraction expansions. We also have that $e$ is not a quadratic irrational, because those have (eventually) periodic continued fraction expansions.
If there were to be a relation between the CF and the transcendence of e,one would feel that it would involve the pattern in the values of the CF. Considering also there appears to be no decipherable pattern to the CF of pi it may well be that such a path does not exist, as if it did then it seems likely that a similar relation would be present in pi.
I thought that all the algebraic transcendentals had a repeated pattern in their continued fraction representation. I know this is true for all the quadratic surds.
The fact that the representation for e, as well as e^2^n don't truly repeat, in the conventional sense, would suggest that they are not algebraic, hence transcendental. Many 'nonalgebraic' numbers, like e and pi, have simple representations as continued fractions.

3$\begingroup$ What is an "algebraic transcendental"? Do you mean "algebraic irrational"? Anyway, what you know about quadratic surds is essentially unique to them: an eventually periodic continued fraction is a quadratic surd. There is no known pattern to the entries in the continued fraction of the cube root of 2. $\endgroup$ – KConrad Feb 24 '11 at 5:04

1$\begingroup$ In what way is the continued fraction representation of pi simple? $\endgroup$ – S. Carnahan♦ Aug 25 '11 at 4:15

$\begingroup$ @S. Carnahan: He might be referring not to the regular continued fraction for $\pi$, but others such as those given here: en.wikipedia.org/wiki/… $\endgroup$ – Todd Trimble♦ Oct 19 '11 at 10:38