Continued fraction with prime reciprocal entries

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.

• @Michael I think what he means is that the continued fraction $[a_1; a_2, a_3, \ldots]$ converges iff the series of partial denominators $\sum_{i=1}^\infty a_i$ diverges. Obviously this implies, as a special case, that $[a_1; a_2, a_3, \ldots]$ converges whenever the $a_i$ are all positive integers. – MJD Sep 3 '13 at 16:06
• For future reference, the convergence theorem mentioned above is Theorem 10 of Continued Fractions 3 ed., A. Ya. Khinchin, pp.10-12. – MJD Sep 5 '13 at 0:48
• Dear Tony: have you tried an inverse symbolic calculator such as isc.carma.newcastle.edu.au ? – Bruno Joyal Oct 21 '13 at 16:09
• 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
• @GTonyJacobs Between 1.142833307 and 1.510380475? Well.... that narrows it down! :) – Bruno Joyal Oct 27 '13 at 23:34