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What is the limit of the continued fraction $$\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{7+\cfrac{1}{11+\cfrac{1}{13+\cdots}}}}}}\ ?$$

Is the limit algebraic, or expressible in terms of e or $\pi$? What is the fastest way to approximate the limit?

Michael Hardy
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Angela Pretorius
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  • Does the sequence $2,3,5,7,11,13,\dots$ consist of all prime numbers? – Américo Tavares Sep 12 '11 at 11:26
  • @Angela Richardson: Are you sure that it is convergent? Also, can you really use the expression of continued fractions to express a complex number such as $i$? Maybe I am missing something here, but I am a little confused. Thanks for any explanation. – awllower Sep 12 '11 at 11:29
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    @awllower: it converges by [Sleszynski-Pringsheim](http://books.google.com/books?id=zUUgs4BgcgoC&pg=PA129). – J. M. ain't a mathematician Sep 12 '11 at 11:40
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    @awllower: It is convergent by the Śleszyńsky-Pringsheim's Theorem. The continued fraction $\mathbf{K}(a_{n}/b_{n})$ converges if for all $n$ $$\left\vert b_{n}\right\vert \geq \left\vert a_{n}\right\vert +1.$$ – Américo Tavares Sep 12 '11 at 11:40
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    Reference: Chapter I, Theorem 4.1 of *Continued fractions with applications* by Lisa Lorentzen and Haakon Waadeland. – Américo Tavares Sep 12 '11 at 11:46
  • It's pretty easy to see that it converges without any theorems whose proof is not obvious. – Michael Hardy Sep 12 '11 at 14:38
  • Thanks for all the explanations. – awllower Sep 14 '11 at 10:19
  • +1 I was curious myself and did a little math with the primes < 100,000. Excel gives 0.4323320871859030 which is off by the last two digits. Then I found this question. – John Alexiou Feb 24 '14 at 21:43
  • Has anything new been found about this continued fraction since this post was last active? – Shaun Apr 15 '14 at 10:31

1 Answers1

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As I've noted in the comments, the Śleszyński-Pringsheim Theorem guarantees that this continued fraction converges to a finite value (since the prime numbers are always greater than or equal to $2$).

This arXiv preprint gives an approximate value $s$ for the continued fraction whose partial denominators are prime numbers:

$$s=0.43233208718590286890925379324199996370511089688\dots$$

which the author computed with PARI/GP. He notes that the Plouffe inverter does not at all recognize this constant.


Here's some Mathematica code for computing the constant to prec or so significant figures:

prec = 500;
y = N[2, prec + 5]; c = y; d = 0; k = 2;
While[True,
  p = Prime[k];
  c = p + 1/c; d = 1/(p + d);
  h = c*d; y *= h;
  If[Abs[h - 1] <= 10^(-prec), Break[]];
  k++];
N[1/y, prec]

which yields the result 0.43233208718590286890925379324199996370511089687765131032815206715855390511529588664247730234675307312901358874751711021925473474173059981681532525370102846860319246045704466728602248840679362020193843643798792955246786129609763893526940277522319731978458635595794036202066338633654544895108909659715862787332585763686200183679952128087865043794610126643260422526400822552675221511335417037835319471839444535578027072057047898703982387228841680143293913635134426277200532815721973910424896503203478507

J. M. ain't a mathematician
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