%I #20 Nov 07 2022 20:27:21

%S 0,9,4,4,4,1,8,5,8,1,9,6,5,0,4,9,4,2,1,8,4

%N Decimal expansion of Sum_{k>=1} k/prime(k)^4.

%C From _Jon E. Schoenfield_, Nov 07 2022: (Start)

%C Let M = 10^10, and let J be the number of primes < M, i.e., J = pi(M) = 455052511; then prime(J+1) = 10000000019.

%C Since prime(J+1) > M+2 and prime(k+1) - prime(k) >= 2 for all k > 1, it follows that, for all k > J,

%C prime(k) > M + 2*(k - J)

%C and thus

%C k/prime(k)^4 < k/(M + 2*(k - J))^4

%C so

%C Sum_{k>J} k/prime(k)^4 < Sum_{k>J} k/(M + 2*(k - J))^4

%C and it can be shown that the sum on the right-hand side is a value < 5*10^-22.

%C Summing the values of k/prime(k)^4 for all k <= J to obtain

%C Sum_{k=1..J} k/prime(k)^4 = 0.0944418581965049421841...

%C yields a lower bound on the infinite sum, and since the infinite sum is

%C Sum_{k>=1} k/prime(k)^4 = Sum_{k=1..J} k/prime(k)^4 + Sum_{k>J} k/prime(k)^4,

%C it must be less than

%C Sum_{k=1..J} k/prime(k)^4 + Sum_{k>J} k/(M + 2*(k - J))^4,

%C which is less than

%C 0.0944418581965049421842 + 5*10^-22 = 0.0944418581965049421847,

%C which thus provides an upper bound on the infinite sum. (End)

%e 0.094441858196504942184...

%Y Cf. A097906, A097879.

%K more,nonn,cons

%O 0,2

%A _Pierre CAMI_, Sep 02 2004

%E a(15)-a(17) corrected and a(18)-a(21) added by _Jon E. Schoenfield_, Nov 07 2022