A continued fraction involving prime numbers

Question

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?

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
@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
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
+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

J. M. ain't a mathematician · Accepted Answer · 2011-09-16T18:02:54.990

31

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

edited Sep 16 '11 at 18:02

answered Sep 12 '11 at 11:47

J. M. ain't a mathematician

75,051

1

For purposes of computation, you can use Lentz-Thompson-Barnett. – J. M. ain't a mathematician Sep 12 '11 at 11:51
4

The OEIS has it, though: http://oeis.org/A084255 – Charles Sep 12 '11 at 12:53
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@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 – Shivam Patel Feb 19 '14 at 16:23
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@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. – J. M. ain't a mathematician May 01 '15 at 13:58
@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. – milia Nov 14 '22 at 19:12
@milia this is proof that the number is irrational. if it were rational, it would have a finite continued fraction representation, but the primes are not finite, so the continued fraction representation is not finite, so the number is irrational – wyboo May 07 '23 at 23:08
@wyboo thanks, that's a good argument for the specific continued series. But what about φ ? Its continued fraction representation doesn't contain prime numbers, since 1 isn't considered a prime number. How do we know that (1;1,1,1,1,...) doesn't stop after many billions of ones proving in effect rationality of φ ? – milia May 11 '23 at 04:10

A continued fraction involving prime numbers

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