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An infinite simple continued fraction

$$\frac{1}{b_1 + \frac{1}{b_2 + \frac{1}{b_3+\dots}}} (b_i\in\mathbb Z\setminus\left\{0\right\})$$

is irrational. Now for a generalized continued fraction:

$$\frac{a_1}{b_1 + \frac{a_2}{b_2 + \frac{a_3}{b_3+\dots}}} (a_i\in\mathbb Z\setminus\left\{0\right\}, b_i\in\mathbb Z),$$

the same conclusion is apparently not valid. Legendre gave a sufficient condition for irrationality:

$$|a_i|<|b_i|$$

for any $i$ large enough.

Can this result be strengthened in any way? Especially, might it hold for $a_i, b_i\in\mathbb Q$?

Also, can anyone give an example of an infinite generalized continued fraction that converges to a rational, showing that some condition on the $a_i, b_i$ is needed for irrationality?

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    $\begingroup$ For the simple continued fraction, the $b_i$ should be positive integers. $\endgroup$
    – Gerry Myerson
    Commented Feb 23, 2017 at 11:15
  • $\begingroup$ Perhaps this is interesting: some heuristical examples and systematized at go.helms-net.de/math/divers/GenContFracRationalE.htm $\endgroup$
    – Gottfried Helms
    Commented Feb 25, 2017 at 12:49
  • $\begingroup$ A better reference is likely David Angell - A family of continued fractions (2010) Journal of Number Theory 130 , pg. 904-911 (Elsevier). $\endgroup$
    – Gottfried Helms
    Commented Feb 25, 2017 at 14:17
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    $\begingroup$ Do you have a source for Legendre's condition applying when $a_i$ or $b_i$ are negative integers? $\endgroup$
    – Pazzaz
    Commented Dec 25, 2019 at 12:19
  • $\begingroup$ It should be worth noting that any continued fraction is quadratic irrational iff it's periodic, see: Wolfram MathWorld $\endgroup$
    – Vessel
    Commented Mar 28 at 19:11

3 Answers 3

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There are such examples in Ramanujan's Notebooks, Part 2, page 116enter image description here

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There is a family of generalized continued fractions which give irrational or rational values depending on a very simple but nontrivial parametrization. I'd like to use the notation $$c = a_0 + {b_0\over a_1 + {b_1\over a_2 + ... }} = a + {b\over (a+1) + {(b+1)\over (a+2) + {(b+2)\over (a+3) + ... }... }}$$ in accordance with the Angell-article (see below).

The following family of generalized continued fraction gives rational or irrational numbers depending on parametrization: $$\begin{array} {r|r} \begin{matrix}&a_k=\\b_k=&\end{matrix} &&0+k & 1+k & 2+k & 3+k & \cdots \\ \hline 0+k && 0 & 1 & 2 &3 & \cdots \\ \hline 1+k && {1\over e-1} & {1\over e-2} & {1\over 2e-5} & {1\over 6e-16} & \cdots \\ \hline 2+k && {1\over 1} & -{1e-1\over 0e-1} & -{1e-2\over 1e-3}& -{2e-5\over 4e-11} & \cdots \\ \hline 3+k && {4\over 3} & {2 \over 1} & -{0e-2\over 1e-2}& -{2e-6\over 3e-8} & \cdots \\ \hline 4+k && {21\over 13} & {9 \over 4} & {3 \over 1} & -{3e-6\over 2e-6} & \cdots \\ \hline \cdots & & \cdots \end{array}$$

I've found that heuristically, using wolframalpha for support in the evaluation.

David Angell describes that family and gives that heuristic the analytic background (at least for the rational results if I got this correctly) see: David Angell - A family of continued fractions (2010) Journal of Number Theory 130 , pg. 904-911 (Elsevier), online: "paywall"

A somewhat larger table is at my mathpages - GenContFrac

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Apparently no such extension is known. But there are other irrationality criteria for sequences of the form

$$\frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i},$$

such as those by Brun and by Froda.

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