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?

  • $\begingroup$ Does the sequence $2,3,5,7,11,13,\dots$ consist of all prime numbers? $\endgroup$ Sep 12, 2011 at 11:26
  • $\begingroup$ @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. $\endgroup$
    – awllower
    Sep 12, 2011 at 11:29
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    $\begingroup$ @awllower: it converges by Sleszynski-Pringsheim. $\endgroup$ Sep 12, 2011 at 11:40
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    $\begingroup$ @awllower: It is convergent by the Śleszyńsky-Pringsheim's Theorem. The continued fraction $\mathbf{K}(a_{n}/b_{n})$ converges if for all $n$ $$\left\vert b_{n}\right\vert \geq \left\vert a_{n}\right\vert +1.$$ $\endgroup$ Sep 12, 2011 at 11:40
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    $\begingroup$ Reference: Chapter I, Theorem 4.1 of Continued fractions with applications by Lisa Lorentzen and Haakon Waadeland. $\endgroup$ Sep 12, 2011 at 11:46

1 Answer 1


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:


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

which yields the result 0.43233208718590286890925379324199996370511089687765131032815206715855390511529588664247730234675307312901358874751711021925473474173059981681532525370102846860319246045704466728602248840679362020193843643798792955246786129609763893526940277522319731978458635595794036202066338633654544895108909659715862787332585763686200183679952128087865043794610126643260422526400822552675221511335417037835319471839444535578027072057047898703982387228841680143293913635134426277200532815721973910424896503203478507

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    $\begingroup$ For purposes of computation, you can use Lentz-Thompson-Barnett. $\endgroup$ Sep 12, 2011 at 11:51
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    $\begingroup$ The OEIS has it, though: oeis.org/A084255 $\endgroup$
    – Charles
    Sep 12, 2011 at 12:53
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    $\begingroup$ @J.M. can you calculate this for twin prime pairs instead of the prime (To a certain extent)...as its irrationality will prove the twin prime conjecture $\endgroup$ Feb 19, 2014 at 16:23
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    $\begingroup$ @Shivam, one can certainly compute a numerical approximation of the constant you want to a pile of digits, but proving irrationality with it just seems rather unlikely. $\endgroup$ May 1, 2015 at 13:58
  • $\begingroup$ @J.M.ain'tamathematician Since the continued fraction goes on to infinity, given that prime numbers are infinite, isn't this an indication that the number is irrational, like φ = (1;1,1,1,1,1,...), for example ? Not sure if this constitutes a proof though. $\endgroup$
    – milia
    Nov 14, 2022 at 19:12

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