%I #23 Nov 01 2020 08:11:32
%S 1,2,4,0,4,0,7,1,4,6,6,5,5,9,6,0,6,2,8,9,4,6,4,1,8,0,2,1,4,0,5,7,2,8,
%T 3,3,9,2,3,1,3,8,1,0,7,3,4,6,9,0,9,9,2,6,9,0,3,7,1,6,4,2,6,1,5,7,4,3,
%U 0,0,2,2,7,5,6,2,1,2,7,2,3,9,6,4,4,7,4,0,1,9
%N Decimal expansion of the sum of reciprocals of the smallest primes > 2^k for k >= 0.
%C 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.
%D 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...
%F Equals Sum_{k>=0} 1/A014210(k).
%e 1.2404071466559606289464180214057283392313810734691...
%p evalf(sum(1/nextprime(2^k), k=0..infinity),90);
%t ndigits = 90; RealDigits[Sum[1/NextPrime[2^k], {k, 0, ndigits/Log10[2] + 1}], 10, ndigits][[1]] (* _Amiram Eldar_, Oct 29 2020 *)
%o (PARI) suminf(k=0, 1/nextprime(2^k+1)) \\ _Michel Marcus_, Oct 29 2020
%Y Cf. A014210, A104080, A203074.
%K nonn,cons
%O 1,2
%A _Bernard Schott_, Oct 29 2020