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.

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    $\begingroup$ For future reference, the convergence theorem mentioned above is Theorem 10 of Continued Fractions 3 ed., A. Ya. Khinchin, pp.10-12. $\endgroup$ – MJD Sep 5 '13 at 0:48
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    $\begingroup$ Dear Tony: have you tried an inverse symbolic calculator such as isc.carma.newcastle.edu.au ? $\endgroup$ – Bruno Joyal Oct 21 '13 at 16:09
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    $\begingroup$ 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. $\endgroup$ – G Tony Jacobs Oct 22 '13 at 16:22
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    $\begingroup$ @GTonyJacobs Between 1.142833307 and 1.510380475? Well.... that narrows it down! :) $\endgroup$ – Bruno Joyal Oct 27 '13 at 23:34
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    $\begingroup$ @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. $\endgroup$ – Kirill Oct 29 '13 at 6:21

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