OFFSET
1,2
COMMENTS
If q(k) = A014210(k) is the smallest prime > 2^k, then 2^k < q(k), so Sum_{k>=0} 1/q(k) < Sum_{k>=0} 1/2^k = 2; hence, the sum of the reciprocals of these primes q(k) form a convergent series.
REFERENCES
J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 615 pp. 82 and 279, Ellipses, Paris, 2004. Warning : gives Sum_{k>=1} 1/A104080(k) = 0.7404...
FORMULA
Equals Sum_{k>=0} 1/A014210(k).
EXAMPLE
1.2404071466559606289464180214057283392313810734691...
MAPLE
evalf(sum(1/nextprime(2^k), k=0..infinity), 90);
MATHEMATICA
ndigits = 90; RealDigits[Sum[1/NextPrime[2^k], {k, 0, ndigits/Log10[2] + 1}], 10, ndigits][[1]] (* Amiram Eldar, Oct 29 2020 *)
PROG
(PARI) suminf(k=0, 1/nextprime(2^k+1)) \\ Michel Marcus, Oct 29 2020
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Bernard Schott, Oct 29 2020
STATUS
approved