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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    $\begingroup$ 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. $\endgroup$ Commented Sep 3, 2013 at 15:49
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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
    Commented Sep 5, 2013 at 0:48
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    $\begingroup$ Dear Tony: have you tried an inverse symbolic calculator such as isc.carma.newcastle.edu.au ? $\endgroup$ Commented Oct 21, 2013 at 16:09
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    $\begingroup$ @GTonyJacobs Between 1.142833307 and 1.510380475? Well.... that narrows it down! :) $\endgroup$ Commented Oct 27, 2013 at 23:34
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    $\begingroup$ Also for reference, the result stated is known as the Seidel-Stern theorem. $\endgroup$
    – TheSimpliFire
    Commented Nov 14, 2020 at 10:50


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