We know that the reciprocals of the primes form a divergent series. We also know that a necessary and sufficient condition for a continued fraction to converge is that its entries diverge as a series. This leads me to wonder to what value this continued fraction converges:

$$\tfrac{1}{2}+\cfrac1{\frac13 + \cfrac1{\frac15 + \cfrac1{\frac17 + \cfrac1{\frac1{11} + \cfrac1{\frac1{13} + \cdots }}}}} $$

I've already shown that the terms of the harmonic series, when placed in a continued fraction, do something nice, namely: $\left[1, 1, \tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{4},\ldots\right] = \frac{\pi}{2}$. This prime reciprocal problem seems harder, though.

Any thoughts are greatly appreciated.

G Tony Jacobs
  • 29,851
  • 4
  • 46
  • 100
  • Just to be clear, you are asking about $$\cfrac{1}{\frac12+\cfrac{1}{\frac13+\cfrac{1}{\frac15+\frac{1}{\frac17+\cdots}}}}?$$ – MJD Sep 03 '13 at 15:47
  • Honestly, it could be that, or just the denominator of that (which is really what I meant), or maybe $1+$ that, because sticking a $1$ on the front made the harmonic one get pretty. I don't mean to be ambiguous about it, but answering any of those would answer the others. What's important is what happens in the tail, not up front. – G Tony Jacobs Sep 03 '13 at 15:49
  • Something that sometimes works is to calculate the numerical value and then look it up on the Internet, say by doing a Google search or by putting the digit sequence into OEIS. If you haven't tried that, you might. This expression converges pretty slowly, so it might take a while to get enough digits though. – MJD Sep 03 '13 at 16:25
  • 2
    For future reference, the convergence theorem mentioned above is Theorem 10 of *Continued Fractions* 3 ed., A. Ya. Khinchin, pp.10-12. – MJD Sep 05 '13 at 00:48
  • 2
    Dear Tony: have you tried an inverse symbolic calculator such as http://isc.carma.newcastle.edu.au/ ? – Bruno Joyal Oct 21 '13 at 16:09
  • 1
    Well, the trouble is that the convergence is very, very slow. I just checked, and using the first 10,000 primes, I can only say that the limit is between 1.142833307 and 1.510380475. There are a lot of numbers on that interval. – G Tony Jacobs Oct 22 '13 at 16:22
  • Using the first 10 million buys you one more digit of precision, at least in binary. I've got upper and lower bounds using convergents obtained from the first 10^k and the first 10^k+1 primes respectively, for k=1 to 7, and I'm thinking of fitting curves to those to predict their limit. I'll post the data here in a few hours. – G Tony Jacobs Oct 25 '13 at 12:45
  • 2
    @GTonyJacobs Between 1.142833307 and 1.510380475? Well.... that narrows it down! :) – Bruno Joyal Oct 27 '13 at 23:34
  • Ok, in the following triples $(k,x_k,y_k)$, we have $x_k$ and $y_k$ are the $10^k$th and the $10^k+1$st convergent respectively: (1, 2.113130611, 0.865979468), (2, 1.706114684, 1.023529212), (3, 1.574693807, 1.099595278), (4, 1.510380426, 1.142836239), (5, 1.471974493, 1.170851435), (6, 1.446299153, 1.190588778), (7, 1.427831121, 1.205300864). Further data points would require considerable computer time to calculate, at least the way I'm doing it. – G Tony Jacobs Oct 28 '13 at 16:19
  • @GTonyJacobs With $10^{11}$ primes, taking the averages of successive convergents, and using some convergence acceleration technique, the number appears to be around $1.31(1)$. – Kirill Oct 28 '13 at 17:23
  • @Kirill, can you say more about how you did that? What program did you use, and what's a "convergence acceleration technique"? – G Tony Jacobs Oct 29 '13 at 04:57
  • 1
    @GTonyJacobs You know $(C_n,C_{n+1})$ bracket the true value, so make a guess that $x_n \approx \frac{C_n+C_{n+1}}{2}$ is a decent approximation (I don't know theory for that, though). After computing $x_n$ for $n=10^4,\ldots,10^{11}$, and plotting, note that it appears as if $$x_n = x_\infty + \frac{\gamma_1}{\log n} + \frac{\gamma_2}{(\log n)^2} $$ holds approximately. Estimate $x_\infty, \gamma_1, \gamma_2$ using least squares, and use $x_\infty$ as your best guess for the true value; magnitudes of $\gamma_i$ should be small. No guarantees, of course, but it probably works. – Kirill Oct 29 '13 at 06:21
  • 1
    @GTonyJacobs I wrote a program to compute this myself, in C++, and used [primesieve](https://code.google.com/p/primesieve/) to generate primes. Using a carefully written prime-sieving library only matters when dealing with billions of primes, though, and you can do well even writing everything yourself. Using a fast compiled language is a must for getting to large numbers. The convergence acceleration technique I used is called Levin transformation (and is slightly different from what I described above). – Kirill Oct 29 '13 at 06:28
  • @GTonyJacobs, how did you prove the value for the harmonics? It appears to be true numerically, but the convergence is extremely slow. Also, was this value known before? – Yuriy S Mar 20 '16 at 10:27
  • 1
    @YuriyS, term-by-term, the convergents are the same as the partial products of Wallis' product. This can be shown by induction. I haven't seen this result, with the harmonic continued fraction, anywhere else, but since it's really just a variation on Wallis' product, it doesn't seem to me to be all that exciting. – G Tony Jacobs Mar 20 '16 at 14:07
  • 1
    @GTonyJacobs, it's on Wiki page about continued fractions. Still, a very interesting result, thank you – Yuriy S Mar 20 '16 at 15:18
  • 1
    By approximating the "tail" of the continued fraction using the [1;1,1/2,1/3,1/4,1/5...] expression for pi/2 (by unwinding it back to get [0;$1/p_n$,$1/(p_n+1)$,$1/(p_n+2)$,...]), the convergence can be vastly improved, and with about 1000 terms, you can get the approximate value of $1.310728$, give or take about $3\times 10^{-7}$. – Glen O Oct 17 '19 at 02:55
  • 1
    As it turned out, because the reverse calculation from the continued fraction for $\pi/2$ rapidly approaches 1, it is easy enough to simply take the approximation as ending in 1, so you'd have [0;1/2,1/3,...,$1/p_n$,1]. This dramatically accelerates the convergence, and considering the results to the prime 99999989 as well as the previous few, it appears that we can confidently place the value as being in the vicinity of 1.31072836773768. This digit sequence does not appear in OEIS. – Glen O Oct 19 '19 at 05:19
  • 1
    Also for reference, the result stated is known as the Seidel-Stern theorem. – TheSimpliFire Nov 14 '20 at 10:50
  • Could you explain how you used the Wallis product to get that $ \frac{\pi}{2}=\left[1, 1, \tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{4},\ldots\right] $? I'm unable to see how it goes. – Dabed Dec 09 '20 at 22:52
  • 1
    @DanielD., it's been a while, but as I recall from beginning to reconstruct it, I used two separate induction arguments. The sequence of partial products of Wallis and the sequence of partial quotients of that continued fraction are identical. It's not too complicated (as I recall) to make an induction argument for the odd-numbered terms of that sequence, and another (very similar) one for the even-numbered terms. If you'd like more details, let me know. Although I can't guarantee speed right now, I'll do what I can :) – G Tony Jacobs Dec 19 '20 at 09:26
  • Thank you very much, If you want I can create a question so that you could post the argument if you feel like it, I didn't make a question because this result looked like classic result since it contains the terms of the harmonic series so I thought it would appear everywhere but I only found one [proof](https://pdfs.semanticscholar.org/a89b/436eb04bec45b1d913ce61c3ef4655bea051.pdf) and I had a hard time finding it, even more I don't get too far before getting lost so if you think there is a simpler way it would be cool. – Dabed Dec 20 '20 at 02:54
  • 1
    @DanielD., if you post this as a question, I'll write up my solution. A new question would be more conducive than this comment section, for sure, and someone else might even come along with a better way than mine! – G Tony Jacobs Dec 20 '20 at 17:13
  • I have finally asked a [question](https://math.stackexchange.com/questions/3956635/how-to-prove-that-frac-pi2-left1-1-tfrac12-tfrac13-tfrac1) I should have done it before instead of bloating with comments here sorry, thanks in advance for your post and feel free also to edit anything about the question – Dabed Dec 20 '20 at 22:28

0 Answers0