Is there a definition of $ 1 + \frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{\ddots}}}}$? I am somewhat familiar with continued fractions; that is, I am aware that their convergence depends on whether our input is rational or not. This is not a homework problem, I am just curious about this topic which has gotten little attention in my current studies. If this expression is defined, what is it and does it converge?
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What other methods do you know that could perhaps check for convergence? – cygorx Oct 21 '13 at 02:40
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In general, I am aware of methods such as the ratio test, n-th root test, and comparison test, but I am not sure how I would apply them here. – Jamil_V Oct 21 '13 at 02:44
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6All simple continued fractions, including yours, converge. This is an immediate corollary of Theorem 10 (p.10) in A. Ya. Khinchin Continued Fractions, University of Chicago Press, 1964. – MJD Oct 21 '13 at 02:46
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It is a simple continued fraction because all the numerators are 1, right? – Jamil_V Oct 21 '13 at 02:47
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That, and because all the partial denominators (in this case 1, 2, 3, etc.) are positive integers. As for the exact value, I would guess offhand that little is known about it; but it is quite easy to see that it is very nearly equal to $\frac{83120346}{57999271} \approx 1.43312742672231\ldots$, so you might try searching for 1.43312742672 and see if you find anything. – MJD Oct 21 '13 at 02:48
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2For example, OEIS has a section about it. To my surprise and delight, there is a closed form! – MJD Oct 21 '13 at 02:50
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Alan Baker proved it transcendental? – Alan Oct 21 '13 at 04:06
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Let $A = [a_0; a_1, a_2, \ldots]$ be shorthand for the continued fraction $$a_0 + \cfrac1{a_1 + \cfrac1{a_2 + \cdots}}.$$ It is perhaps the most important theorem of continued fractions that $A$ is always well-defined when the $a_i$ are positive integers. This is an immediate consequence of a more general theorem which states that $A$ converges if and only if the series $$\sum_{i=0}^\infty a_i$$ diverges. (See Theorem 10 (p. 10) in A. Ya. Khinchin, Continued Fractions, University of Chicago Press 1964.)
Since the $a_i$ in this case increase fairly quickly, the partial quotients of the continued fraction itself converge fairly quickly, and a computer calculation of the truncated fraction $[1; 2,3,4,5,6,7,8,9,10,11]$ yields the approximation $$A\approx 1.43312742672231,$$ which is good to 14 decimal places. (I used the cf-evaluate program found here.) Plugging the sequence of digits into OEIS reveals that this is OEIS sequence A060997 and the claim that the value is exactly $$\frac{I_0(2)}{I_1(2)}$$ where $I_0$ and $I_1$ are modified Bessel functions of the first kind. (Unfortunately, OEIS gives neither a proof nor a reference.)
OEIS also claims (again unfortunately without proof; I suppose it follows from the Taylor series expansions for the Bessel functions) that $$A = \frac{\sum_{n=0}^\infty \frac{1}{n!^2}}{\sum_{n=0}^\infty \frac{n}{n!^2}}.$$
It does at least provide an email address for the author of the entry, so you could write and ask him for a reference.
[ Addendum 2016-08-20: Jack D'Aurizio explains the Bessel function thing in https://math.stackexchange.com/a/1871798 .]
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2A reference for the material involving the ratio of Bessel functions and the continued fraction constant: "Transcendental Numbers" Carl Ludwig Siegel (1949) pp. 65-72 – Alan Oct 21 '13 at 19:08