# 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.

• 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. Commented Sep 3, 2013 at 15:49
• For future reference, the convergence theorem mentioned above is Theorem 10 of Continued Fractions 3 ed., A. Ya. Khinchin, pp.10-12.
– MJD
Commented Sep 5, 2013 at 0:48
• Dear Tony: have you tried an inverse symbolic calculator such as isc.carma.newcastle.edu.au ? Commented Oct 21, 2013 at 16:09
• @GTonyJacobs Between 1.142833307 and 1.510380475? Well.... that narrows it down! :) Commented Oct 27, 2013 at 23:34
• Also for reference, the result stated is known as the Seidel-Stern theorem. Commented Nov 14, 2020 at 10:50